Download Summer Assignment ACP Pre-calculus Do as many problems as

Summer Assignment
ACP Pre-calculus
Do as many problems as you are able to do
P1 Algebraic expressions, mathematical models, and real numbers
Page 15-17
Questions 11-39 every other odd, 85-101 odd
P2 Exponents and scientific notation
Page 28
Questions 5-
5 by 5’s
P4 Polynomials
Page 53
Questions 1-81 odd
P7 Equations
Linear, absolute value, quadratic
Page 981
Questions 1-15 odd, 43-54 all, 55-109 odd
Prerequisites:
Fundamental
Concepts of Algebra
P
What can algebra
possibly have to
tell me about
• the skyrocketing
cost of a college
education?
• my workouts?
• apologizing to a friend
for a blunder I committed?
• the meaning of the
national debt that exceeds
$9 trillion?
• time dilation on a futuristic
high-speed journey to a nearby
star?
• the widening imbalance between
numbers of women and men on
college campuses?
This chapter reviews fundamental
concepts of algebra that are prerequisites
for the study of precalculus. Throughout the
chapter, you will see how the special language
of algebra describes your world.
Here’s where you’ll find these applications:
• College costs: Section P.1, Example 2; Exercise
Set P.1, Exercises 131–132
• Workouts: Exercise Set P.1, Exercises 129–130
• Apologizing: Essay on page 15
• The national debt: Section P.2, Example 6
• Time dilation: Essay on page 42
• College gender imbalance: Chapter P Test, Exercise 32.
1
2 Chapter P Prerequisites: Fundamental Concepts of Algebra
Section
P.1
Objectives
Algebraic Expressions, Mathematical Models,
and Real Numbers
H
� Evaluate algebraic
ow would your lifestyle change if a gallon of gas
cost $9.15? Or if the price of a staple such as milk
was $15? That’s how much those products would cost if
their prices had increased at the same rate college
tuition has increased since 1980. (Source: Center for
College Affordability and Productivity) In this section,
you will learn how the special language of algebra
describes your world, including the skyrocketing cost of
a college education.
expressions.
� Use mathematical models.
� Find the intersection of two
sets.
� Find the union of two sets.
� Recognize subsets of the
real numbers.
� Use inequality symbols.
� Evaluate absolute value.
� Use absolute value to
express distance.
� Identify properties of the
real numbers.
쐅 Simplify algebraic
expressions.
Algebraic Expressions
Algebra uses letters, such as x and y, to represent numbers.
If a letter is used to represent various numbers, it is called a
variable. For example, imagine that you are basking in the sun on the beach. We
can let x represent the number of minutes that you can stay in the sun without
burning with no sunscreen. With a number 6 sunscreen, exposure time without
burning is six times as long, or 6 times x. This can be written 6 # x, but it is usually
expressed as 6x. Placing a number and a letter next to one another indicates
multiplication.
Notice that 6x combines the number 6 and the variable x using the operation
of multiplication. A combination of variables and numbers using the operations of
addition, subtraction, multiplication, or division, as well as powers or roots, is called
an algebraic expression. Here are some examples of algebraic expressions:
x + 6,฀ x - 6,฀ 6x,฀
x
,฀ 3x + 5,฀ x2 - 3,฀ 1x + 7.
6
Many algebraic expressions involve exponents. For example, the algebraic
expression
17x 2 + 261x + 3257
approximates the average cost of tuition and fees at public U.S. colleges for
the school year ending x years after 2000. The expression x2 means x # x, and is
read “x to the second power” or “x squared.” The exponent, 2, indicates that the
base, x, appears as a factor two times.
Exponential Notation
If n is a counting number (1, 2, 3, and so on),
Exponent or Power
bn= b ⴢ b ⴢ b ⴢ . . . ⴢ b.
Base
b appears as a
factor n times.
bn is read “the nth power of b” or “b to the nth power.” Thus, the nth power of b
is defined as the product of n factors of b. The expression bn is called an
exponential expression. Furthermore, b1 = b.
For example,
82 = 8 # 8 = 64,฀ 53 = 5 # 5 # 5 = 125,฀ and฀ 2 4 = 2 # 2 # 2 # 2 = 16.
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 3
�
Evaluate algebraic expressions.
Evaluating Algebraic Expressions
Evaluating an algebraic expression means to find the value of the expression for a
given value of the variable.
Many algebraic expressions involve more than one operation. Evaluating an
algebraic expression without a calculator involves carefully applying the following
order of operations agreement:
The Order of Operations Agreement
1. Perform operations within the innermost parentheses and work outward.
If the algebraic expression involves a fraction, treat the numerator and the
denominator as if they were each enclosed in parentheses.
2. Evaluate all exponential expressions.
3. Perform multiplications and divisions as they occur, working from left to right.
4. Perform additions and subtractions as they occur, working from left to right.
Evaluating an Algebraic Expression
EXAMPLE 1
Evaluate 7 + 51x - 423 for x = 6.
Solution
7 + 51x - 423 = 7 + 516 - 423
= 7 + 51223
= 7 + 5182
= 7 + 40
= 47
Check Point
�
Use mathematical models.
1
Replace x with 6.
First work inside parentheses: 6 - 4 = 2.
Evaluate the exponential expression:
23 = 2 # 2 # 2 = 8.
Multiply: 5182 = 40.
Add.
Evaluate 8 + 61x - 322 for x = 13.
Formulas and Mathematical Models
An equation is formed when an equal sign is placed between two algebraic
expressions. One aim of algebra is to provide a compact, symbolic description of the
world. These descriptions involve the use of formulas. A formula is an equation that
uses variables to express a relationship between two or more quantities.
Here are two examples of formulas related to heart rate and exercise.
Couch-Potato Exercise
1
H= (220-a)
5
Heart rate, in
beats per minute,
is
1 of
5
the difference between
220 and your age.
Working It
9
H= (220-a)
10
Heart rate, in
beats per minute,
is
9 of
10
the difference between
220 and your age.
4 Chapter P Prerequisites: Fundamental Concepts of Algebra
The process of finding formulas to describe real-world phenomena is called
mathematical modeling. Such formulas, together with the meaning assigned to the
variables, are called mathematical models. We often say that these formulas model,
or describe, the relationships among the variables.
EXAMPLE 2
Modeling the Cost of Attending a Public College
The bar graph in Figure P.1 shows the average cost of tuition and fees for public
four-year colleges, adjusted for inflation.The formula
T = 17x 2 + 261x + 3257
models the average cost of tuition and fees, T, for public U.S. colleges for the school
year ending x years after 2000.
a. Use the formula to find the average cost of tuition and fees at public U.S. colleges
for the school year ending in 2007.
b. By how much does the formula underestimate or overestimate the actual cost
shown in Figure P.1?
Average Cost of Tuition and Fees at Public
Four-Year United States Colleges
$6200
5836
$5800
5491
Tuition and Fees
$5400
5132
$5000
4694
$4600
4081
$4200
3725
$3800
$3400
3362
3487
$3000
2000 2001 2002 2003 2004 2005 2006 2007
Ending Year in the School Year
Figure P.1
Source: The College Board
Solution
a. Because 2007 is 7 years after 2000, we substitute 7 for x in the given formula.
Then we use the order of operations to find T, the average cost of tuition and
fees for the school year ending in 2007.
T = 17x2 + 261x + 3257
This is the given mathematical model.
T = 171722 + 261172 + 3257
Replace each occurrence of x with 7.
T = 171492 + 261172 + 3257
Evaluate the exponential expression 72 = 7 # 7 = 49.
T = 833 + 1827 + 3257
Multiply from left to right: 171492 = 833 and
261172 = 1827.
T = 5917
Add.
The formula indicates that for the school year ending in 2007, the average cost
of tuition and fees at public U.S. colleges was $5917.
b. Figure P.1 shows that the average cost of tuition and fees for the school year
ending in 2007 was $5836.
The cost obtained from the formula, $5917, overestimates the actual data value
by $5917 - $5836, or by $81.
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 5
2 Assuming trends indicated by the data in Figure P.1 continue,
use the formula T = 17x2 + 261x + 3257, described in Example 2, to project the
average cost of tuition and fees at public U.S. colleges for the school year ending
in 2010.
Check Point
Sometimes a mathematical model gives an estimate that is not a good approximation or is extended to include values of the variable that do not make sense. In
these cases, we say that model breakdown has occurred. For example, it is not likely
that the formula in Example 2 would give a good estimate of tuition and fees in 2050
because it is too far in the future. Thus, model breakdown would occur.
Sets
Before we describe the set of real numbers, let’s be sure you are familiar with some
basic ideas about sets. A set is a collection of objects whose contents can be clearly
determined. The objects in a set are called the elements of the set. For example, the
set of numbers used for counting can be represented by
51, 2, 3, 4, 5, Á 6.
Study Tip
Grouping symbols such as parentheses,
( ), and square brackets, [ ], are not
used to represent sets. Only commas
are used to separate the elements of
a set. Separators such as colons or
semicolons are not used.
The braces, 5฀ 6, indicate that we are representing a set. This form of representation,
called the roster method, uses commas to separate the elements of the set. The
symbol consisting of three dots after the 5, called an ellipsis, indicates that there is
no final element and that the listing goes on forever.
A set can also be written in set-builder notation. In this notation, the elements
of the set are described, but not listed. Here is an example:
{x|x is a counting number less than 6}.
The set of all x
such that
x is a counting number less than 6.
The same set written using the roster method is
�
Find the intersection of two sets.
51, 2, 3, 4, 56.
If A and B are sets, we can form a new set consisting of all elements that are in
both A and B. This set is called the intersection of the two sets.
Definition of the Intersection of Sets
A
B
The intersection of sets A and B, written A ¨ B, is the set of elements common
to both set A and set B. This definition can be expressed in set-builder notation
as follows:
A ¨ B = 5x ƒ x is an element of A AND x is an element of B6.
A傽B
Figure P.2 Picturing the
intersection of two sets
Figure P.2 shows a useful way of picturing the intersection of sets A and B. The
figure indicates that A ¨ B contains those elements that belong to both A and B at
the same time.
Finding the Intersection of Two Sets
EXAMPLE 3
Find the intersection: 57, 8, 9, 10, 116 ¨ 56, 8, 10, 126.
Solution The elements common to 57, 8, 9, 10, 116 and 56, 8, 10, 126 are 8 and
10. Thus,
57, 8, 9, 10, 116 ¨ 56, 8, 10, 126 = 58, 106.
Check Point
3
Find the intersection: 53, 4, 5, 6, 76 ¨ 53, 7, 8, 96.
6 Chapter P Prerequisites: Fundamental Concepts of Algebra
If a set has no elements, it is called the empty set, or the null set, and is
represented by the symbol ¤ (the Greek letter phi). Here is an example that shows
how the empty set can result when finding the intersection of two sets:
{2, 4, 6} 艚 {3, 5, 7}=¤.
These sets have no
common elements.
�
Find the union of two sets.
Their intersection
has no elements
and is the empty set.
Another set that we can form from sets A and B consists of elements that are
in A or B or in both sets. This set is called the union of the two sets.
Definition of the Union of Sets
A
B
A艛B
The union of sets A and B, written A ´ B, is the set of elements that are
members of set A or of set B or of both sets. This definition can be expressed in
set-builder notation as follows:
A ´ B = 5x ƒ x is an element of A OR x is an element of B6.
Figure P.3 Picturing the union
of two sets
Study Tip
When finding the union of two sets,
do not list twice any elements that
appear in both sets.
Figure P.3 shows a useful way of picturing the union of sets A and B. The
figure indicates that A ´ B is formed by joining the sets together.
We can find the union of set A and set B by listing the elements of set A. Then,
we include any elements of set B that have not already been listed. Enclose all
elements that are listed with braces. This shows that the union of two sets is also a set.
Finding the Union of Two Sets
EXAMPLE 4
Find the union: 57, 8, 9, 10, 116 ´ 56, 8, 10, 126.
Solution To find 57, 8, 9, 10, 116 ´ 56, 8, 10, 126, start by listing all the elements
from the first set, namely 7, 8, 9, 10, and 11. Now list all the elements from the second
set that are not in the first set, namely 6 and 12. The union is the set consisting of all
these elements. Thus,
57, 8, 9, 10, 116 ´ 56, 8, 10, 126 = 56, 7, 8, 9, 10, 11, 126.
Check Point
�
Recognize subsets of the real
numbers.
4
Find the union: 53, 4, 5, 6, 76 ´ 53, 7, 8, 96.
The Set of Real Numbers
The sets that make up the real numbers are summarized in Table P.1. We refer to
these sets as subsets of the real numbers, meaning that all elements in each subset
are also elements in the set of real numbers.
Notice the use of the symbol L in the examples of irrational numbers. The
symbol means “is approximately equal to.” Thus,
22 L 1.414214.
Technology
A calculator with a square root key
gives a decimal approximation for
22, not the exact value.
We can verify that this is only an approximation by multiplying 1.414214 by itself.
The product is very close to, but not exactly, 2:
1.414214 * 1.414214 = 2.000001237796.
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 7
Table P.1 Important Subsets of the Real Numbers
Name
Description
Natural
numbers
⺞
{1, 2, 3, 4, 5, . . . }
These are the numbers that we use for
counting.
2, 3, 5, 17
Whole
numbers
⺧
{0, 1, 2, 3, 4, 5, . . . }
The set of whole numbers includes 0 and
the natural numbers.
0, 2, 3, 5, 17
Integers
⺪
{. . . , –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, . . .}
The set of integers includes the negatives of
the natural numbers and the whole numbers.
–17, –5, –3, –2, 0, 2, 3, 5, 17
{ ba | a and b are integers and b ⫽ 0}
–17= 1 , –5= 1 , –3, –2,
0, 2, 3, 5, 17,
Rational
numbers
⺡
This means that b is not equal to zero.
The set of rational numbers is the set of all
numbers that can be expressed as a quotient
of two integers, with the denominator not 0.
Rational numbers can be expressed as
terminating or repeating decimals.
Irrational
numbers
⺙
Real numbers
Rational
numbers
Integers
Irrational
numbers
Whole
numbers
Natural
numbers
Figure P.4 Every real number
is either rational or irrational.
Examples
The set of irrational numbers is the set of
all numbers whose decimal representations
are neither terminating nor repeating.
Irrational numbers cannot be expressed as a
quotient of integers.
–17
2
=0.4,
5
–2
=–0.6666
3
–5
. . . =–0.6
兹2≠1.414214
–兹3≠–1.73205
p≠3.142
–
p
≠–1.571
2
Not all square roots are irrational. For example, 225 = 5 because
52 = 5 # 5 = 25. Thus, 225 is a natural number, a whole number, an integer, and a
rational number A 225 = 51 B .
The set of real numbers is formed by taking the union of the sets of rational
numbers and irrational numbers. Thus, every real number is either rational or
irrational, as shown in Figure P.4.
Real Numbers
The set of real numbers is the set of numbers that are either rational or
irrational:
5x ƒ x is rational or x is irrational6.
The symbol ⺢ is used to represent the set of real numbers. Thus,
⺢ = 5x ƒ x is rational6 ´ 5x ƒ x is irrational6.
EXAMPLE 5
Recognizing Subsets of the Real Numbers
Consider the following set of numbers:
3
e - 7, - , 0, 0.6, 25, p, 7.3, 281 f.
4
List the numbers in the set that are
a. natural numbers.
b. whole numbers.
c. integers.
d. rational numbers.
Solution
e. irrational numbers.
f. real numbers.
a. Natural numbers: The natural numbers are the numbers used for counting. The
only natural number in the set E -7, - 34 , 0, 0.6, 25, p, 7.3, 281 F is 281
because 281 = 9. (9 multiplied by itself, or 92, is 81.)
8 Chapter P Prerequisites: Fundamental Concepts of Algebra
b. Whole numbers: The whole numbers consist of the natural numbers and 0. The
elements of the set E -7, - 34 , 0, 0.6, 25, p, 7.3, 281 F that are whole numbers
are 0 and 281.
c. Integers: The integers consist of the natural numbers, 0, and the negatives of
the natural numbers. The elements of the set E -7, - 34 , 0, 0.6, 25, p, 7.3,
281 F that are integers are 281 , 0, and -7.
d. Rational numbers: All numbers in the set E -7, - 34 , 0, 0.6, 25, p, 7.3, 281 F
that can be expressed as the quotient of integers are rational numbers. These
include - 7 A -7 = -17 B , - 34 , 0 A 0 = 01 B , and 281 A 281 = 91 B . Furthermore, all
numbers in the set that are terminating or repeating decimals are also rational
numbers. These include 0.6 and 7.3.
e. Irrational numbers: The irrational numbers in the set E -7, - 34 , 0, 0.6, 25, p,
7.3, 281 F are 25 A 25 L 2.236 B and p1p L 3.142. Both 25 and p are
only approximately equal to 2.236 and 3.14, respectively. In decimal form, 25
and p neither terminate nor have blocks of repeating digits.
f. Real numbers: All the numbers in the given set E -7, - 34 , 0, 0.6, 25, p, 7.3,
281 F are real numbers.
Check Point
5
Consider the following set of numbers:
e - 9, -1.3, 0, 0.3,
p
, 29, 210 f.
2
List the numbers in the set that are
a. natural numbers.
b. whole numbers.
d. rational numbers. e. irrational numbers.
c. integers.
f. real numbers.
The Real Number Line
The real number line is a graph used to represent the set of real numbers. An
arbitrary point, called the origin, is labeled 0. Select a point to the right of 0 and label
it 1. The distance from 0 to 1 is called the unit distance. Numbers to the right of the
origin are positive and numbers to the left of the origin are negative. The real
number line is shown in Figure P.5.
Negative
direction
−7
−6
−5
−4
−3
−2
−1
0
1
2
Negative numbers
3
4
5
6
7
Positive
direction
Positive numbers
Figure P.5 The real number line
Study Tip
Wondering how we located 22 as a
precise point on the number line in
Figure P.6? We used a right triangle
with both legs of length 1. The
remaining side measures 22 .
1
Rational
numbers
1
兹2
0
Real numbers are graphed on a number line by placing a dot at the correct
location for each number. The integers are easiest to locate. In Figure P.6, we’ve
graphed six rational numbers and three irrational numbers on a real number line.
1
2
兹2
We’ll have lots more to say about
right triangles later in the chapter.
Irrational
numbers
−2
−2
−兹2 ≈ −1.4
− 21
−1
0
0
0.3 =
1
3
1
兹2 ≈฀1.4
11
4
2
=฀2 43
3
兹16 =฀4
4
5
p ≈฀3.14
Figure P.6 Graphing numbers on a real number line
Every real number corresponds to a point on the number line and every point
on the number line corresponds to a real number. We say that there is a one-to-one
correspondence between all the real numbers and all points on a real number line.
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers 9
�
Use inequality symbols.
Ordering the Real Numbers
On the real number line, the real numbers increase from left to right. The lesser of
two real numbers is the one farther to the left on a number line. The greater of two
real numbers is the one farther to the right on a number line.
Look at the number line in Figure P.7. The integers - 4 and - 1 are graphed.
−5
−4
−3
−2
−1
0
1
2
3
4
5
Figure P.7
Observe that - 4 is to the left of -1 on the number line. This means that -4 is
less than -1.
−4 is less than −1 because −4 is to
the left of −1 on the number line.
–4<–1
In Figure P.7, we can also observe that - 1 is to the right of - 4 on the number
line. This means that - 1 is greater than -4.
–1>–4
−1 is greater than −4 because −1 is to
the right of −4 on the number line.
The symbols 6 and 7 are called inequality symbols. These symbols always
point to the lesser of the two real numbers when the inequality statement is true.
−4 is less than −1.
−1 is greater than −4.
–4<–1
The symbol points to -4, the lesser
number.
–1>–4 The symbol still points to -4, the
lesser number.
The symbols 6 and 7 may be combined with an equal sign, as shown in the
following table:
This inequality is true
if either the < part or
the = part is true.
This inequality is true
if either the > part or
the =฀ part is true.
�
Evaluate absolute value.
Symbols
Meaning
Examples
Explanation
aⱕb
a is less than or equal to b.
2ⱕ9
9ⱕ9
Because 2<9
Because 9=9
bⱖa
b is greater than or equal to a.
9ⱖ2
2ⱖ2
Because 9>2
Because 2=2
Absolute Value
The absolute value of a real number a, denoted by ƒ a ƒ , is the distance from 0 to a on
the number line. This distance is always taken to be nonnegative. For example, the
real number line in Figure P.8 shows that
ƒ -3 ƒ = 3฀ and฀ ƒ 5 ƒ = 5.
兩−3兩 = 3
兩5兩 = 5
−5 −4 −3 −2 −1 0 1 2 3 4 5
Figure P.8 Absolute value as the
distance from 0
The absolute value of -3 is 3 because -3 is 3 units from 0 on the number line. The
absolute value of 5 is 5 because 5 is 5 units from 0 on the number line. The absolute
value of a positive real number or 0 is the number itself. The absolute value of a
negative real number, such as -3, is the number without the negative sign.
10 Chapter P Prerequisites: Fundamental Concepts of Algebra
We can define the absolute value of the real number x without referring to a
number line. The algebraic definition of the absolute value of x is given as follows:
Definition of Absolute Value
ƒxƒ = b
x
-x
if x Ú 0
if x 6 0
If x is nonnegative (that is, x Ú 0), the absolute value of x is the number itself.
For example,
ƒ5ƒ = 5
1
1
` ` =
3
3
ƒpƒ = p
兩0兩=0.
Zero is the only number
whose absolute value is 0.
If x is a negative number (that is, x 6 0), the absolute value of x is the opposite of x.
This makes the absolute value positive. For example,
兩–3兩=–(–3)=3
兩–p兩=–(–p)=p
2 – 1 2 =– a– 1 b = 1 .
3
3
3
This middle step is usually omitted.
Evaluating Absolute Value
EXAMPLE 6
Rewrite each expression without absolute value bars:
a. ƒ 23 - 1 ƒ
b. ƒ 2 - p ƒ
c.
ƒxƒ
if x 6 0.
x
Solution
a. Because 23 L 1.7, the number inside the absolute value bars, 23 - 1, is
positive. The absolute value of a positive number is the number itself. Thus,
ƒ 23 - 1 ƒ = 23 - 1.
b. Because p L 3.14, the number inside the absolute value bars, 2 - p, is
negative. The absolute value of x when x 6 0 is -x. Thus,
ƒ 2 - p ƒ = - 12 - p2 = p - 2.
c. If x 6 0, then ƒ x ƒ = - x. Thus,
ƒxƒ
-x
=
= - 1.
x
x
Check Point
a. ƒ 1 - 22 ƒ
6
Rewrite each expression without absolute value bars:
b. ƒ p - 3 ƒ
c.
ƒxƒ
if x 7 0.
x
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers
11
Listed below are several basic properties of absolute value. Each of these
properties can be derived from the definition of absolute value.
Discovery
Properties of Absolute Value
Verify the triangle inequality if a = 4
and b = 5. Verify the triangle
inequality if a = 4 and b = - 5.
When does equality occur in the
triangle inequality and when does
inequality occur? Verify your observation with additional number pairs.
For all real numbers a and b,
1. ƒ a ƒ Ú 0
2. ƒ -a ƒ = ƒ a ƒ
ƒaƒ
a
4. ƒ ab ƒ = ƒ a ƒ ƒ b ƒ
5. ` ` =
,฀ b Z 0
b
ƒbƒ
6. ƒ a + b ƒ … ƒ a ƒ + ƒ b ƒ (called the triangle inequality)
�
Use absolute value to express
distance.
3. a … ƒ a ƒ
Distance between Points on a Real Number Line
Absolute value is used to find the distance between two points on a real number
line. If a and b are any real numbers, the distance between a and b is the absolute
value of their difference. For example, the distance between 4 and 10 is 6. Using
absolute value, we find this distance in one of two ways:
兩10-4兩=兩6兩=6
or
兩4-10兩=兩–6兩=6.
The distance between 4 and 10 on the real number line is 6.
Notice that we obtain the same distance regardless of the order in which we subtract.
Distance between Two Points on the Real Number Line
If a and b are any two points on a real number line, then the distance between a
and b is given by
ƒ a - b ƒ ฀ or฀ ƒ b - a ƒ .
Distance between Two Points on a Number Line
EXAMPLE 7
Find the distance between -5 and 3 on the real number line.
Solution Because the distance between a and b is given by ƒ a - b ƒ , the distance
between -5 and 3 is
兩–5-3兩=兩–8兩=8.
a = −5
8
−5 −4 −3 −2 −1 0 1 2 3 4 5
Figure P.9 verifies that there are 8 units between -5 and 3 on the real number line.
We obtain the same distance if we reverse the order of the subtraction:
ƒ 3 - 1-52 ƒ = ƒ 8 ƒ = 8.
Figure P.9 The distance between
-5 and 3 is 8.
Check Point
�
Identify properties of the real
numbers.
b=3
7
Find the distance between -4 and 5 on the real number line.
Properties of Real Numbers and Algebraic Expressions
When you use your calculator to add two real numbers, you can enter them in any
order. The fact that two real numbers can be added in any order is called the
commutative property of addition. You probably use this property, as well as other
12 Chapter P Prerequisites: Fundamental Concepts of Algebra
properties of real numbers listed in Table P.2, without giving it much thought. The
properties of the real numbers are especially useful when working with algebraic
expressions. For each property listed in Table P.2, a, b, and c represent real numbers,
variables, or algebraic expressions.
Table P.2 Properties of the Real Numbers
Name
The Associative
Property and the
English Language
In the English language, phrases
can take on different meanings
depending on the way the words
are associated with commas.
Here are three examples.
Meaning
Commutative
Property of
Addition
Changing order
when adding does
not affect the sum.
a + b = b + a
Commutative
Property of
Multiplication
Changing order when
multiplying does not
affect the product.
ab = ba
Associative
Property of
Addition
Changing grouping when
adding does not affect
the sum.
1a + b2 + c = a + 1b + c2
Associative
Property of
Multiplication
Changing grouping when
multiplying does not
affect the product.
1ab2c = a1bc2
Distributive
Property of
Multiplication
over Addition
Multiplication distributes
over addition.
Examples
• 13 + 7 = 7 + 13
• 13x + 7 = 7 + 13x
• 22 # 25 = 25 # 22
• x # 6 = 6x
• 3 + 18 + x2 = 13 + 82 + x
= 11 + x
• -213x2 = 1- 2 # 32x = - 6x
• 7 A 4+兹3 B =7 ⴢ 4+7 ⴢ 兹3
=28+7兹3
a ⴢ (b+c)=a ⴢ b+a ⴢ c
• 5(3x+7)=5 ⴢ 3x+5 ⴢ 7
=15x+35
• 23 + 0 = 23
• 0 + 6x = 6x
Identity
Property of
Addition
Zero can be deleted
from a sum.
a + 0 = a
0 + a = a
• What’s the latest dope?
What’s the latest, dope?
Identity
Property of
Multiplication
• Population of Amsterdam
broken down by age and sex
Population of Amsterdam,
broken down by age and sex
One can be deleted
from a product.
a#1 = a
1#a = a
Inverse
Property of
Addition
The sum of a real
number and its additive
inverse gives 0,
the additive identity.
a + 1 - a2 = 0
•
•
•
•
The product of a
nonzero real number
and its multiplicative
inverse gives 1,
the multiplicative identity.
1
a # = 1,฀ a Z 0
a
1#
a = 1,฀ a Z 0
a
• 7#
• Woman, without her man, is
nothing.
Woman, without her, man is
nothing.
1 -a2 + a = 0
Inverse
Property of
Multiplication
• 1#p = p
• 13x # 1 = 13x
25 + A - 25 B = 0
-p + p = 0
6x + 1-6x2 = 0
1-4y2 + 4y = 0
1
= 1
7
1
b1x - 32 = 1,฀ x Z 3
• a
x - 3
The properties of the real numbers in Table P.2 apply to the operations of
addition and multiplication. Subtraction and division are defined in terms of
addition and multiplication.
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers
13
Definitions of Subtraction and Division
Let a and b represent real numbers.
Subtraction: a - b = a + 1-b2
We call -b the additive inverse or opposite of b.
Division: a , b = a # b1 , where b Z 0
We call b1 the multiplicative inverse or reciprocal of b. The quotient of a and
b, a , b, can be written in the form ba , where a is the numerator and b the
denominator of the fraction.
Because subtraction is defined in terms of adding an inverse, the distributive
property can be applied to subtraction:
a(b-c)=ab-ac
(b-c)a=ba-ca.
For example,
4(2x-5)=4 ⴢ 2x-4 ⴢ 5=8x-20.
쐅
Simplify algebraic expressions.
Simplifying Algebraic Expressions
The terms of an algebraic expression are those parts that are separated by addition.
For example, consider the algebraic expression
7x - 9y + z - 3,
which can be expressed as
7x + 1-9y2 + z + 1- 32.
This expression contains four terms, namely 7x, -9y, z, and - 3.
The numerical part of a term is called its coefficient. In the term 7x, the 7 is the
coefficient. If a term containing one or more variables is written without a coefficient,
the coefficient is understood to be 1. Thus, z means 1z. If a term is a constant, its
coefficient is that constant. Thus, the coefficient of the constant term -3 is - 3.
7x+(–9y)+z+(–3)
Coefficient
is 7.
Coefficient
is −9.
Coefficient
is 1; z
means 1z.
Coefficient
is −3.
The parts of each term that are multiplied are called the factors of the term.
The factors of the term 7x are 7 and x.
Like terms are terms that have exactly the same variable factors. For example,
3x and 7x are like terms. The distributive property in the form
ba + ca = 1b + c2a
enables us to add or subtract like terms. For example,
Study Tip
To combine like terms mentally, add
or subtract the coefficients of the
terms. Use this result as the coefficient
of the terms’ variable factor(s).
3x + 7x = 13 + 72x = 10x
7y 2 - y2 = 7y2 - 1y2 = 17 - 12y2 = 6y2.
This process is called combining like terms.
An algebraic expression is simplified when parentheses have been removed
and like terms have been combined.
14 Chapter P Prerequisites: Fundamental Concepts of Algebra
Simplifying an Algebraic Expression
EXAMPLE 8
Simplify:
612x2 + 4x2 + 1014x2 + 3x2.
Solution
6(2x¤+4x)+10(4x¤+3x)
=6 ⴢ 2x¤+6 ⴢ 4x+10 ⴢ 4x¤+10 ⴢ 3x
52x2 and 54x are not like terms.
They contain different variable
factors, x2 and x, and cannot
be combined.
=12x¤+24x+40x¤+30x
=(12x¤+40x¤)+(24x+30x)
=52x¤+54x
8
Check Point
Simplify:
Use the distributive property to
remove the parentheses.
Multiply.
Group like terms.
Combine like terms.
714x2 + 3x2 + 215x2 + x2.
Properties of Negatives
The distributive property can be extended to cover more than two terms within
parentheses. For example,
This sign represents
subtraction.
This sign tells us
that the number is negative.
–3(4x-2y+6)=–3 ⴢ 4x-(–3) ⴢ 2y-3 ⴢ 6
= - 12x - 1-6y2 - 18
= - 12x + 6y - 18.
The voice balloons illustrate that negative signs can appear side by side. They
can represent the operation of subtraction or the fact that a real number is
negative. Here is a list of properties of negatives and how they are applied to
algebraic expressions:
Properties of Negatives
Let a and b represent real numbers, variables, or algebraic expressions.
Property
Examples
1. 1 -12a = - a
1 -124xy = - 4xy
2. -1- a2 = a
3. 1- a2b = - ab
- 1-6y2 = 6y
1-724xy = - 7 # 4xy = - 28xy
4. a1- b2 = - ab
5x1 -3y2 = - 5x # 3y = - 15xy
5. - 1a + b2 = - a - b
-17x + 6y2 = - 7x - 6y
6. - 1a - b2 = - a + b
= b - a
-13x - 7y2 = - 3x + 7y
= 7y - 3x
It is not uncommon to see algebraic expressions with parentheses preceded by
a negative sign or subtraction. Properties 5 and 6 in the box, - 1a + b2 = - a - b
and -1a - b2 = - a + b, are related to this situation. An expression of the form
-1a + b2 can be simplified as follows:
–(a+b)=–1(a+b)=(–1)a+(–1)b=–a+(–b)=–a-b.
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers
15
Do you see a fast way to obtain the simplified expression on the right at the
bottom of the previous page? If a negative sign or a subtraction symbol appears outside parentheses, drop the parentheses and change the sign of every term within the
parentheses. For example,
-13x2 - 7x - 42 = - 3x2 + 7x + 4.
Simplifying an Algebraic Expression
EXAMPLE 9
Simplify:
Solution
8x + 235 - 1x - 324.
8x + 235 - 1x - 324
= 8x + 235 - x + 34
= 8x + 238 - x4
= 8x + 16 - 2x
Drop parentheses and change the sign of each
term in parentheses: - 1x - 32 = - x + 3.
Simplify inside brackets: 5 + 3 = 8.
Apply the distributive property:
278 ⴚ x8 ⴝ 2 ⴢ 8 ⴚ 2x ⴝ 16 ⴚ 2x.
= 18x - 2x2 + 16
= 18 - 22x + 16
= 6x + 16
Check Point
9
Simplify:
Group like terms.
Apply the distributive property.
Simplify.
6 + 437 - 1x - 224.
Using Algebra to Solve Problems in Your Everyday Life
Should You Apologize to Your Friend?
In this formula, each variable is assigned a number from 1 (low) to 10 (high).
How big a deal is the issue?
In Geek Logik (Workman Publishing,
2006), humorist Garth Sundem presents
formulas covering dating, romance, career,
finance, everyday decisions, and health. On
the right is a sample of one of his formulas
that “takes the guesswork out of life,
providing easier living through algebra.”
How peeved
is your friend?
D[Rp(Ra+P)+D(Ra-Rp)]=A
How responsible
are you for this blunder?
How responsible does your friend
perceive you to be in this matter?
If A is 5 or
greater, you
should apologize.
Exercise Set P.1
Practice Exercises
5. x2 + 3x, for x = 8
6. x2 + 5x, for x = 6
In Exercises 1–16, evaluate each algebraic expression for the given
value or values of the variable(s).
7. x2 - 6x + 3, for x = 7
8. x2 - 7x + 4, for x = 8
1. 7 + 5x, for x = 10
2. 8 + 6x, for x = 5
9. 4 + 51x - 723, for x = 9
10. 6 + 51x - 623, for x = 8
3. 6x - y, for x = 3 and y = 8
11. x2 - 31x - y2, for x = 8 and y = 2
4. 8x - y, for x = 3 and y = 4
12. x2 - 41x - y2, for x = 8 and y = 3
16 Chapter P Prerequisites: Fundamental Concepts of Algebra
51x + 22
71x - 32
14.
, for x = 10
, for x = 9
2x - 14
2x - 16
2x + 3y
15.
, for x = - 2 and y = 4
x + 1
2x + y
16.
, for x = - 2 and y = 4
xy - 2x
13.
The formula
C =
5
1F - 322
9
expresses the relationship between Fahrenheit temperature, F, and
Celsius temperature, C. In Exercises 17–18, use the formula to
convert the given Fahrenheit temperature to its equivalent
temperature on the Celsius scale.
17. 50°F
18. 86°F
A football was kicked vertically upward from a height of 4 feet
with an initial speed of 60 feet per second. The formula
h = 4 + 60t - 16t2
In Exercises 51–60, rewrite each expression without absolute
value bars.
51. ƒ 300 ƒ
52. ƒ - 203 ƒ
55. ƒ 22 - 5 ƒ
-3
57.
ƒ -3 ƒ
59. 7 -3 ƒ - ƒ -7 7
54. ƒ 7 - p ƒ
56. ƒ 25 - 13 ƒ
-7
58.
ƒ -7 ƒ
60. 7 - 5 ƒ - ƒ - 137
53. ƒ 12 - p ƒ
In Exercises 61–66, evaluate each algebraic expression for x = 2
and y = - 5.
61. ƒ x + y ƒ
62. ƒ x - y ƒ
63. ƒ x ƒ + ƒ y ƒ
64. ƒ x ƒ - ƒ y ƒ
65.
y
66.
ƒyƒ
ƒyƒ
ƒxƒ
+
x
y
In Exercises 67–74, express the distance between the given
numbers using absolute value. Then find the distance by evaluating
the absolute value expression.
describes the ball’s height above the ground, h, in feet, t seconds
after it was kicked. Use this formula to solve Exercises 19–20.
67. 2 and 17
68. 4 and 15
69. - 2 and 5
70. - 6 and 8
19. What was the ball’s height 2 seconds after it was kicked?
71. - 19 and - 4
72. - 26 and - 3
20. What was the ball’s height 3 seconds after it was kicked?
73. -3.6 and -1.4
74. -5.4 and -1.2
In Exercises 21–28, find the intersection of the sets.
21. 51, 2, 3, 46 ¨ 52, 4, 56
23. 5s, e, t6 ¨ 5t, e, s6
25. 51, 3, 5, 76 ¨ 52, 4, 6, 8, 106
26. 50, 1, 3, 56 ¨ 5 -5, -3, - 16
27. 5a, b, c, d6 ¨ ⭋
22. 51, 3, 76 ¨ 52, 3, 86
24. 5r, e, a, l6 ¨ 5l, e, a, r6
28. 5w, y, z6 ¨ ⭋
In Exercises 29–34, find the union of the sets.
29. 51, 2, 3, 46 ´ 52, 4, 56
30. 51, 3, 7, 86 ´ 52, 3, 86
33. 5a, e, i, o, u6 ´ ⭋
34. 5e, m, p, t, y6 ´ ⭋
31. 51, 3, 5, 76 ´ 52, 4, 6, 8, 106
32. 50, 1, 3, 56 ´ 52, 4, 66
In Exercises 35–38, list all numbers from the given set that are
a. natural numbers, b. whole numbers, c. integers, d. rational numbers,
e. irrational numbers, f. real numbers.
35.
36.
37.
38.
E - 9, - 54 , 0, 0.25, 23 , 9.2, 2100 F
E - 7, -0.6, 0, 249, 250 F
E - 11, - 65 , 0, 0.75, 25 , p, 264 F
E - 5, -0.3, 0, 22, 24 F
39. Give an example of a whole number that is not a natural
number.
40. Give an example of a rational number that is not an integer.
41. Give an example of a number that is an integer, a whole
number, and a natural number.
42. Give an example of a number that is a rational number, an
integer, and a real number.
Determine whether each statement in Exercises 43–50 is true or false.
43. - 13 … - 2
44. - 6 7 2
45. 4 Ú - 7
46. - 13 6 - 5
47. - p Ú - p
48. - 3 7 - 13
49. 0 Ú - 6
50. 0 Ú - 13
In Exercises 75–84, state the name of the property illustrated.
75. 6 + 1 -42 = 1-42 + 6
76. 11 # 17 + 42 = 11 # 7 + 11 # 4
77. 6 + 12 + 72 = 16 + 22 + 7
78. 6 # 12 # 32 = 6 # 13 # 22
79. 12 + 32 + 14 + 52 = 14 + 52 + 12 + 32
80. 7 # 111 # 82 = 111 # 82 # 7
81. 21- 8 + 62 = - 16 + 12
82. -813 + 112 = - 24 + 1-882
83.
1
1x + 32 = 1, x Z - 3
1x + 32
84. 1x + 42 + 3- 1x + 424 = 0
In Exercises 85–96, simplify each algebraic expression.
85. 513x + 42 - 4
86. 215x + 42 - 3
87. 513x - 22 + 12x
88. 215x - 12 + 14x
89. 713y - 52 + 214y + 32
90. 412y - 62 + 315y + 102
93. 7 - 433 - 14y - 524
94. 6 - 538 - 12y - 424
91. 513y - 22 - 17y + 22
92. 415y - 32 - 16y + 32
95. 18x + 4 - 361x - 22 + 54
2
2
96. 14x2 + 5 - 371x2 - 22 + 44
In Exercises 97–102, write each algebraic expression without
parentheses.
97. -1- 14x2
99. -12x - 3y - 62
101. 1313x2 + 314y2 + 1- 4y24
102. 1212y2 + 31 -7x2 + 7x4
98. - 1- 17y2
100. -15x - 13y - 12
Section P.1 Algebraic Expressions, Mathematical Models, and Real Numbers
Practice Plus
In Exercises 103–110, insert either 6, 7 , or = in the shaded area
to make a true statement.
103. ƒ - 6 ƒ
ƒ -3 ƒ
3
105. ` `
5
ƒ -0.6 ƒ
5
106. ` `
2
14 # 15
15 14
3
30
107.
40
4
109.
104. ƒ - 20 ƒ
8
8
,
13
13
ƒ - 50 ƒ
ƒ - 2.5 ƒ
17 # 18
108.
18 17
ƒ -1 ƒ
5
50
60
6
4
4
,
17
17
110. ƒ - 2 ƒ
limits of these ranges are fractions of the maximum heart rate,
220 - a. Exercises 129–130 are based on the information in the
graph.
Target Heart Rate Ranges for Exercise Goals
Exercise Goal
Boost performance
as a competitive athlete
Improve cardiovascular
conditioning
Lose weight
Improve overall health and
reduce risk of heart attack
W
In Exercises 111–120, use the order of operations to simplify each
expression.
111. 82 - 16 , 2 2 # 4 - 3
113.
112. 102 - 100 , 52 # 2 - 3
5 # 2 - 32
332 - 1-2242
114.
10 , 2 + 3 # 4
112 - 3 # 222
q
118.
5 - 8
15 - 62 - 2 ƒ 3 - 7 ƒ
2
119.
89 - 3 # 52
120.
Lower limit of range
Upper limit of range
61 -42 - 51-32
Lower limit of range
Upper limit of range
122. A number decreased by the difference between eight and
the number
7
(220-a)
10
4
H= (220-a).
5
H=
1
(220-a)
2
3
H= (220-a).
5
H=
a. What is the lower limit of the heart range, in beats per
minute, for a 30-year-old with this exercise goal?
123. Six times the product of negative five and a number
124. Ten times the product of negative four and a number
126. The difference between the product of six and a number
and negative two times the number
b. What is the upper limit of the heart range, in beats per
minute, for a 30-year-old with this exercise goal?
The bar graph shows the average cost of tuition and fees at private
four-year colleges in the United States.
Average Cost of Tuition and Fees at Private
Four-Year United States Colleges
127. The difference between eight times a number and six more
than three times the number
$23,000
128. Eight decreased by three times the sum of a number and six
22,218
$22,000
21,235
The maximum heart rate, in beats per minute, that you should
achieve during exercise is 220 minus your age:
220-a.
Tuition and Fees
$21,000
Application Exercises
The bar graph at the top of the next column shows the target heart
rate ranges for four types of exercise goals. The lower and upper
19,710
$20,000
$19,000
20,082
18,273
$18,000
17,272
$17,000
$16,000
This algebraic expression gives maximum
heart rate in terms of age, a.
1
130. If your exercise goal is to improve overall health, the graph
shows the following range for target heart rate, H, in beats
per minute:
121. A number decreased by the sum of the number and four
125. The difference between the product of five and a number
and twice the number
Ï
b. What is the upper limit of the heart range, in beats per
minute, for a 20-year-old with this exercise goal?
12 , 3 # 5 ƒ 2 2 + 32 ƒ
In Exercises 121–128, write each English phrase as an algebraic
expression. Then simplify the expression. Let x represent the
number.
R
a. What is the lower limit of the heart range, in beats per
minute, for a 20-year-old with this exercise goal?
9 - 10
7 + 3 - 62
Î
129. If your exercise goal is to improve cardiovascular conditioning,
the graph shows the following range for target heart rate, H, in
beats per minute:
116. 8 - 33- 215 - 72 - 514 - 224
21- 22 - 41-32
E
Fraction of Maximum Heart Rate, 220 − a
115. 8 - 33-212 - 52 - 418 - 624
117.
17
16,233
15,518
$15,000
2000 2001 2002 2003 2004 2005 2006 2007
Ending Year in the School Year
Source: The College Board
18 Chapter P Prerequisites: Fundamental Concepts of Algebra
The formula
T = 15,395 + 988x - 2x
2
models the average cost of tuition and fees, T, at private U.S.
colleges for the school year ending x years after 2000. Use this
information to solve Exercises 131–132.
131. a. Use the formula to find the average cost of tuition
and fees at private U.S. colleges for the school year
ending in 2007.
b. By how much does the formula underestimate or
overestimate the actual cost shown by the graph at the
bottom of the previous page for the school year ending
in 2007?
c. Use the formula to project the average cost of tuition
and fees at private U.S. colleges for the school year
ending in 2010.
132. a. Use the formula to find the average cost of tuition
and fees at private U.S. colleges for the school year
ending in 2006.
b. By how much does the formula underestimate or
overestimate the actual cost shown by the graph at the
bottom of the previous page for the school year ending
in 2006?
c. Use the formula to project the average cost of tuition
and fees at private U.S. colleges for the school year
ending in 2012.
133. You had $10,000 to invest. You put x dollars in a safe,
government-insured certificate of deposit paying 5% per
year. You invested the remainder of the money in
noninsured corporate bonds paying 12% per year. Your
total interest earned at the end of the year is given by the
algebraic expression
0.05x + 0.12110,000 - x2.
a. Simplify the algebraic expression.
b. Use each form of the algebraic expression to determine
your total interest earned at the end of the year if you
invested $6000 in the safe, government-insured
certificate of deposit.
134. It takes you 50 minutes to get to campus.You spend t minutes
walking to the bus stop and the rest of the time riding the bus.
Your walking rate is 0.06 mile per minute and the bus travels
at a rate of 0.5 mile per minute. The total distance walking
and traveling by bus is given by the algebraic expression
136. If n is a natural number, what does bn mean? Give an
example with your explanation.
137. What does it mean when we say that a formula models realworld phenomena?
138. What is the intersection of sets A and B?
139. What is the union of sets A and B?
140. How do the whole numbers differ from the natural numbers?
141. Can a real number be both rational and irrational? Explain
your answer.
142. If you are given two real numbers, explain how to
determine which is the lesser.
143. Think of a situation where you either apologized or did not
apologize for a blunder you committed. Use the formula in
the essay on page 15 to determine whether or not you
should have apologized. How accurately does the formula
model what you actually did?
144. Read Geek Logik by Garth Sundem (Workman Publishing,
2006). Would you recommend the book to college algebra
students? Were you amused by the humor? Which formulas
did you find most useful? Apply at least one of the formulas
by plugging your life data into the equation, using the order
of operations, and coming up with an answer to one of life’s
dilemmas.
Critical Thinking Exercises
Make Sense? In Exercises 145–148, determine whether each
statement makes sense or does not make sense, and explain your
reasoning.
145. My mathematical model describes the data for tuition and
fees at public four-year colleges for the past ten years
extremely well, so it will serve as an accurate prediction for
the cost of public colleges in 2050.
146. A model that describes the average cost of tuition and fees
at private U.S. colleges for the school year ending x years
after 2000 cannot be used to estimate the cost of private
education for the school year ending in 2000.
147. The humor in this cartoon is based on the fact that the
football will never be hiked.
0.06t + 0.5150 - t2.
a. Simplify the algebraic expression.
b. Use each form of the algebraic expression to determine the total distance that you travel if you spend
20 minutes walking to the bus stop.
Writing in Mathematics
Writing about mathematics will help you learn mathematics. For
all writing exercises in this book, use complete sentences to
respond to the question. Some writing exercises can be answered in
a sentence; others require a paragraph or two. You can decide how
much you need to write as long as your writing clearly and directly
answers the question in the exercise. Standard references such as a
dictionary and a thesaurus should be helpful.
135. What is an algebraic expression? Give an example with
your explanation.
FOXTROT © 2000 Bill Amend. Reprinted with permission of UNIVERSAL
PRESS SYNDICATE. All rights reserved.
148. Just as the commutative properties change groupings, the
associative properties change order.
In Exercises 149–156, determine whether each statement is true or
false. If the statement is false, make the necessary change(s) to
produce a true statement.
149. Every rational number is an integer.
150. Some whole numbers are not integers.
151. Some rational numbers are not positive.
Section P.2 Exponents and Scientific Notation 19
b. b5 # b5 = 1b # b # b # b # b21b # b # b # b # b2 = b?
152. Irrational numbers cannot be negative.
153. The term x has no coefficient.
c. Generalizing from parts (a) and (b), what should be
done with the exponents when multiplying exponential
expressions with the same base?
154. 5 + 31x - 42 = 81x - 42 = 8x - 32
155. -x - x = - x + 1-x2 = 0
156. x - 0.021x + 2002 = 0.98x - 4
161. In parts (a) and (b), complete each statement.
In Exercises 157–159, insert either 6 or 7 in the shaded area
between the numbers to make the statement true.
157. 22
3.14
159. 2
p
2
Preview Exercises
160. In parts (a) and (b), complete each statement.
a. b4 # b3 = 1b # b # b # b21b # b # b2 = b?
P.2
b.
b#b#b#b#b#b#b#b
b8
= b?
=
2
b#b
b
c. Generalizing from parts (a) and (b), what should be
done with the exponents when dividing exponential
expressions with the same base?
Exercises 160–162 will help you prepare for the material covered
in the next section.
Section
b#b#b#b#b#b#b
b7
= b?
=
3
b#b#b
b
- 3.5
158. - p
1.5
a.
162. If 6.2 is multiplied by 103, what does this multiplication do
to the decimal point in 6.2?
Exponents and Scientific Notation
Objectives
� Use properties of exponents.
� Simplify exponential
expressions.
� Use scientific notation.
�
Use properties of exponents.
L
istening to the radio on the
way to campus, you hear
politicians
discussing
the
problem of the national debt,
which exceeds $9 trillion. They
state that it’s more than the
gross domestic product of
China, the world’s secondrichest nation, and four times
greater than the combined net
worth of America’s 691 billionaires. They make it seem
like the national debt is a real problem, but later you realize that you don’t really
know what a number like 9 trillion means. If the national debt were evenly divided
among all citizens of the country, how much would every man, woman, and child
have to pay? Is economic doomsday about to arrive?
In this section, you will learn to use exponents to provide a way of putting
large and small numbers in perspective. Using this skill, we will explore the meaning
of the national debt.
Properties of Exponents
The major properties of exponents are summarized in the box that follows and
continues on the next page.
Properties of Exponents
Property
Study Tip
The Negative-Exponent Rule
When a negative integer appears as
an exponent, switch the position of
the base (from numerator to denominator or from denominator to
numerator) and make the exponent
positive.
If b is any real number other than
0 and n is a natural number, then
1
b-n = n .
b
Examples
• 5-3 =
•
1
1
=
3
125
5
1
1
=
= 42 = 16
-2
1
4
42
Section P.2 Exponents and Scientific Notation 19
b. b5 # b5 = 1b # b # b # b # b21b # b # b # b # b2 = b?
152. Irrational numbers cannot be negative.
153. The term x has no coefficient.
c. Generalizing from parts (a) and (b), what should be
done with the exponents when multiplying exponential
expressions with the same base?
154. 5 + 31x - 42 = 81x - 42 = 8x - 32
155. -x - x = - x + 1-x2 = 0
156. x - 0.021x + 2002 = 0.98x - 4
161. In parts (a) and (b), complete each statement.
In Exercises 157–159, insert either 6 or 7 in the shaded area
between the numbers to make the statement true.
157. 22
3.14
159. 2
p
2
Preview Exercises
160. In parts (a) and (b), complete each statement.
a. b4 # b3 = 1b # b # b # b21b # b # b2 = b?
P.2
b.
b#b#b#b#b#b#b#b
b8
= b?
=
2
b#b
b
c. Generalizing from parts (a) and (b), what should be
done with the exponents when dividing exponential
expressions with the same base?
Exercises 160–162 will help you prepare for the material covered
in the next section.
Section
b#b#b#b#b#b#b
b7
= b?
=
3
b#b#b
b
- 3.5
158. - p
1.5
a.
162. If 6.2 is multiplied by 103, what does this multiplication do
to the decimal point in 6.2?
Exponents and Scientific Notation
Objectives
� Use properties of exponents.
� Simplify exponential
expressions.
� Use scientific notation.
�
Use properties of exponents.
L
istening to the radio on the
way to campus, you hear
politicians
discussing
the
problem of the national debt,
which exceeds $9 trillion. They
state that it’s more than the
gross domestic product of
China, the world’s secondrichest nation, and four times
greater than the combined net
worth of America’s 691 billionaires. They make it seem
like the national debt is a real problem, but later you realize that you don’t really
know what a number like 9 trillion means. If the national debt were evenly divided
among all citizens of the country, how much would every man, woman, and child
have to pay? Is economic doomsday about to arrive?
In this section, you will learn to use exponents to provide a way of putting
large and small numbers in perspective. Using this skill, we will explore the meaning
of the national debt.
Properties of Exponents
The major properties of exponents are summarized in the box that follows and
continues on the next page.
Properties of Exponents
Property
Study Tip
The Negative-Exponent Rule
When a negative integer appears as
an exponent, switch the position of
the base (from numerator to denominator or from denominator to
numerator) and make the exponent
positive.
If b is any real number other than
0 and n is a natural number, then
1
b-n = n .
b
Examples
• 5-3 =
•
1
1
=
3
125
5
1
1
=
= 42 = 16
-2
1
4
42
20 Chapter P Prerequisites: Fundamental Concepts of Algebra
The Zero-Exponent Rule
If b is any real number other than 0,
• 70 = 1
• 1-520 = 1
b0 = 1.
•
–50=–1
Only 5 is raised to
the zero power.
The Product Rule
If b is a real number or algebraic
expression, and m and n are integers,
b
m
#b
n
m+n
= b
• 22 # 23 = 22 + 3 = 25 = 32
• x-3 # x7 = x-3 + 7 = x4
.
When multiplying exponential expressions with the same base, add the
exponents. Use this sum as the exponent of the common base.
The Power Rule
If b is a real number or algebraic
expression, and m and n are integers,
1bm2 = bmn.
n
#
3
• 1222 = 22 3 = 26 = 64
1
#
4
• 1x-32 = x-3 4 = x-12 = 12
x
When an exponential expression is raised to a power, multiply the
exponents. Place the product of the exponents on the base and remove
the parentheses.
Study Tip
The Quotient Rule
45
represent different
43
4
numbers:
If b is a nonzero real number or algebraic
expression, and m and n are integers,
•
28
= 28 - 4 = 24 = 16
24
bm
= bm - n.
bn
•
1
x3
= x3 - 7 = x-4 = 4
7
x
x
43
5
and
4
3
45
= 43 - 5 = 4-2 =
1
1
=
16
42
45
= 45 - 3 = 42 = 16.
43
When dividing exponential expressions with the same nonzero base, subtract the
exponent in the denominator from the exponent in the numerator. Use this
difference as the exponent of the common base.
Products Raised to Powers
If a and b are real numbers or algebraic
expressions, and n is an integer,
1ab2n = anbn.
• 1-2y24 = 1 -224 y4 = 16y4
• 1-2xy23 = 1 -223 x3 y3 = - 8x3 y3
When a product is raised to a power, raise each factor to that power.
Quotients Raised to Powers
If a and b are real numbers, b Z 0, or
algebraic expressions, and n is an integer,
a n
an
a b = n.
b
b
24
16
2 4
• a b = 4 =
5
625
5
1- 323
27
3 3
= - 3
• a- b =
x
x3
x
When a quotient is raised to a power, raise the numerator to that power and
divide by the denominator to that power.
Section P.2 Exponents and Scientific Notation 21
�
Simplify exponential
expressions.
Simplifying Exponential Expressions
Properties of exponents are used to simplify exponential expressions. An exponential
expression is simplified when
• No parentheses appear.
• No powers are raised to powers.
• Each base occurs only once.
• No negative or zero exponents appear.
Simplifying Exponential Expressions
Example
1. If necessary, remove parentheses by using
a n
an
1ab2n = anbn฀ or฀ a b = n .
b
b
1xy23 = x3y3
2. If necessary, simplify powers to powers by using
1bm2 = bmn.
#
3
1x42 = x4 3 = x12
n
3. If necessary, be sure that each base appears only
once by using
bm
bm # bn = bm + n฀ or฀ n = bm - n.
b
4. If necessary, rewrite exponential expressions
with zero powers as 1 1b0 = 12. Furthermore,
write the answer with positive exponents by using
1
1
b-n = n ฀ or฀ -n = bn.
b
b
x4 # x3 = x4 + 3 = x7
1
x5
= x5 - 8 = x -3 = 3
8
x
x
The following example shows how to simplify exponential expressions.
Throughout the example, assume that no variable in a denominator is equal to zero.
EXAMPLE 1
Simplifying Exponential Expressions
Simplify:
a. 1- 3x y 2
4 5 3
b. 1-7xy 21 - 2x y 2
4
5 6
c.
- 35x2 y4
5x6y -8
-3
4x2
d. ¢
≤ .
y
Solution
a. 1-3x4 y52 = 1-3231x42 1y52
3
3
3
# #
= 1-323 x4 3 y5 3
= - 27x12y15
Raise each factor inside the parentheses to the
third power.
Multiply the exponents when raising powers to powers.
1-323 = 1-321 -321 -32 = - 27
b. 1-7xy421 - 2x5y 62 = 1 -721 -22xx5 y4y6
Group factors with the same base.
= 14x1 + 5y4 + 6
When multiplying expressions with the
same base, add the exponents.
= 14x6y10
Simplify.
22 Chapter P Prerequisites: Fundamental Concepts of Algebra
-35x2y4
c.
= a
6 -8
5x y
y4
-35 x2
b ¢ 6 ≤ ¢ -8 ≤
5
x
y
= - 7x2 - 6y4 - 1-82
4x2
≤
y
When dividing expressions with the same base,
subtract the exponents.
= - 7x -4 y12
Simplify. Notice that 4 - 1-82 = 4 + 8 = 12.
-7y12
Write as a fraction and move the base with the
negative exponent, x-4, to the other side of the
fraction bar and make the negative exponent
positive.
=
d. ¢
Group factors with the same base.
x4
=
14x22
=
4 -31x22
-3
=
=
=
Check Point
a. 12x3y62
4
-3
Raise the numerator and the denominator to the -3 power.
y -3
-3
y -3
4 -3 x -6
y -3
Multiply the exponents when raising a power to a power:
1x22-3 = x21-32 = x-6.
y3
Move each base with a negative exponent to the other side of
the fraction bar and make each negative exponent positive.
3 6
4 x
y3
43 = 4 # 4 # 4 = 64
64x6
1
Raise each factor in the numerator to the - 3 power.
Simplify:
b. 1 - 6x2y5213xy32
c.
100x 12 y2
20x16 y -4
d. ¢
5x -2
≤ .
y4
Study Tip
Try to avoid the following common errors that can occur when simplifying exponential expressions.
Correct
#b
32 # 34
b
3
4
= b
Incorrect
Description of Error
#b = b
32 # 34 = 96
7
b
= 36
3
4
12
The exponents should be added, not multiplied.
The common base should be retained, not multiplied.
516
= 512
54
516
= 54
54
The exponents should be subtracted, not divided.
14a23 = 64a3
14a23 = 4a3
Both factors should be cubed.
b-n =
1
bn
1a + b2-1 =
b-n = 1
a + b
1
bn
1a + b2-1 =
Only the exponent should change sign.
1
1
+
a
b
The exponent applies to the entire expression a + b.
Section P.2 Exponents and Scientific Notation 23
�
Use scientific notation.
Table P.3 Names of Large
Numbers
102
hundred
10
3
thousand
10
6
million
Scientific Notation
As of December 2007, the national debt of the United States was about $9.2 trillion.
This is the amount of money the government has had to borrow over the years, mostly
by selling bonds, because it has spent more than it has collected in taxes. A stack of
$1 bills equaling the national debt would rise to twice the distance from Earth to the
moon. Because a trillion is 1012 (see Table P.3), the national debt can be expressed as
9.2 * 1012.
The number 9.2 * 1012 is written in a form called scientific notation.
109
billion
1012
trillion
Scientific Notation
quadrillion
A number is written in scientific notation when it is expressed in the form
10
15
1018
quintillion
1021
sextillion
1024
septillion
1027
octillion
10
30
nonillion
10
100
googol
10googol
googolplex
a * 10n,
where the absolute value of a is greater than or equal to 1 and less than 10
11 … ƒ a ƒ 6 102, and n is an integer.
It is customary to use the multiplication symbol, * , rather than a dot, when
writing a number in scientific notation.
Converting from Scientific to Decimal Notation
Here are two examples of numbers in scientific notation:
6.4 * 105฀ means฀ 640,000.
2.17 * 10-3฀ means฀ 0.00217.
Do you see that the number with the positive exponent is relatively large and the
number with the negative exponent is relatively small?
We can use n, the exponent on the 10 in a * 10n, to change a number in
scientific notation to decimal notation. If n is positive, move the decimal point in a
to the right n places. If n is negative, move the decimal point in a to the left
ƒ n ƒ places.
Converting from Scientific to Decimal Notation
EXAMPLE 2
Write each number in decimal notation:
a. 6.2 * 107
b. -6.2 * 107
c. 2.019 * 10 -3
d. - 2.019 * 10-3.
Solution In each case, we use the exponent on the 10 to determine how far to move
the decimal point and in which direction. In parts (a) and (b), the exponent is positive,
so we move the decimal point to the right. In parts (c) and (d), the exponent is negative,
so we move the decimal point to the left.
a. 6.2*10‡=62,000,000
n=7
n=7
Move the decimal point
7 places to the right.
c. 2.019*10–‹=0.002019
n = −3
Check Point
b. –6.2*10‡=–62,000,000
d. –2.019*10–‹=–0.002019
Move the decimal point
兩−3兩 places, or 3 places,
to the left.
2
a. -2.6 * 10
9
Move the decimal point
7 places to the right.
n = −3
Move the decimal point
兩−3兩 places, or 3 places,
to the left.
Write each number in decimal notation:
b. 3.017 * 10-6.
24 Chapter P Prerequisites: Fundamental Concepts of Algebra
Converting from Decimal to Scientific Notation
To convert from decimal notation to scientific notation, we reverse the procedure of
Example 2.
Converting from Decimal to Scientific Notation
Write the number in the form a * 10n.
• Determine a, the numerical factor. Move the decimal point in the given number
to obtain a number whose absolute value is between 1 and 10, including 1.
• Determine n, the exponent on 10n. The absolute value of n is the number of
places the decimal point was moved. The exponent n is positive if the decimal point was moved to the left, negative if the decimal point was moved to
the right, and 0 if the decimal point was not moved.
Converting from Decimal Notation
to Scientific Notation
EXAMPLE 3
Write each number in scientific notation:
b. -34,970,000,000,000
d. -0.0000000000802.
a. 34,970,000,000,000
c. 0.0000000000802
Solution
Technology
a. 34,970,000,000,000=3.497*10⁄‹
You can use your calculator’s 冷EE 冷
(enter exponent) or 冷EXP 冷 key
to convert from decimal to scientific
notation. Here is how it’s done for
0.0000000000802.
Many Scientific Calculators
Keystrokes
Move the decimal point to get
a number whose absolute value
is between 1 and 10.
The decimal point was
moved 13 places to the
left, so n = 13.
b. –34,970,000,000,000=–3.497*10⁄‹
c. 0.0000000000802=8.02*10–⁄⁄
.0000000000802 冷EE 冷 冷 =
Study Tip
If the absolute value of a
number is greater than 10, it
will have a positive exponent in
scientific notation. If the
absolute value of a number is
less than 1, it will have a
negative exponent in scientific
notation.
冷
Display
8.02 - 11
Move the decimal point to get
a number whose absolute value
is between 1 and 10.
The decimal point was
moved 11 places to the
right, so n = −11.
Many Graphing Calculators
d. –0.0000000000802=–8.02*10–⁄⁄
Use the mode setting for scientific
notation.
Check Point
3
Write each number in scientific notation:
Keystrokes
.0000000000802 冷ENTER 冷
a. 5,210,000,000
b. -0.00000006893.
Display
8.02E - 11
EXAMPLE 4
Expressing the U.S. Population in Scientific
Notation
As of December 2007, the population of the United States was approximately
303 million. Express the population in scientific notation.
Solution Because a million is 106, the 2007 population can be expressed as
303*10fl.
This factor is not between 1 and 10, so
the number is not in scientific notation.
Section P.2 Exponents and Scientific Notation 25
The voice balloon indicates that we need to convert 303 to scientific notation.
303*10fl=(3.03*10¤)*10fl=3.03*10¤± fl=3.03*10°
303 = 3.03 ×฀102
In scientific notation, the population is 3.03 * 108.
Check Point
4
Express 410 * 107 in scientific notation.
Computations with Scientific Notation
Properties of exponents are used to perform computations with numbers that are
expressed in scientific notation.
EXAMPLE 5
Computations with Scientific Notation
Perform the indicated computations, writing the answers in scientific notation:
a. 16.1 * 105214 * 10-92
b.
1.8 * 104
.
3 * 10-2
Solution
a. 16.1 * 105214 * 10-92
Technology
16.1 * 105214 * 10-92
On a Calculator:
= 16.1 * 42 * 1105 * 10-92
= 24.4 * 105 + 1-92
Many Scientific Calculators
6.1 冷EE 冷 5 冷 * 冷 4 冷EE 冷 9 冷฀ +/ - ฀ 冷 冷 =
冷
Convert 24.4 to scientific notation:
24.4 = 2.44 * 101.
101 * 10-4 = 101 + 1-42 = 10-3
= 2.44 * 10-3
Many Graphing Calculators
6.1 冷EE 冷 5 冷 * 冷 4 冷EE 冷 冷1-2 冷 9 冷ENTER 冷
2.44E - 3
Simplify.
= 12.44 * 1012 * 10-4
2.44 - 03
Display (in scientific notation mode)
Add the exponents on 10 and multiply the
other parts.
= 24.4 * 10-4
Display
b.
1.8 * 104
1.8
104
= a
b * ¢ -2 ≤
-2
3
3 * 10
10
= 0.6 * 104 - 1-22
= 0.6 * 106
= 16 * 10-12 * 106
= 6 * 105
Check Point
5
Regroup factors.
Regroup factors.
Subtract the exponents on 10 and divide the
other parts.
Simplify: 4 - 1 - 22 = 4 + 2 = 6.
Convert 0.6 to scientific notation: 0.6 = 6 * 10-1.
10-1 * 106 = 10-1 + 6 = 105
Perform the indicated computations, writing the answers in
scientific notation:
a. 17.1 * 105215 * 10-72
b.
1.2 * 106
.
3 * 10-3
26 Chapter P Prerequisites: Fundamental Concepts of Algebra
Applications: Putting Numbers in Perspective
Due to tax cuts and spending increases, the United States began accumulating
large deficits in the 1980s. To finance the deficit, the government had borrowed
$9.2 trillion as of December 2007. The graph in Figure P.10 shows the national
debt increasing over time.
National Debt (trillions of dollars)
The National Debt
Figure P.10
Source: Office of Management and Budget
10
9
8
7
9.2
6
5
4
3
2
1
0
4.9
5.2
5.4
5.5
5.6
5.7
5.8
6.2
6.8
7.4
7.9
8.5
3.2
1.8
0.9
1980 1985 1990 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007
Year
Example 6 shows how we can use scientific notation to comprehend the
meaning of a number such as 9.2 trillion.
EXAMPLE 6
The National Debt
As of December 2007, the national debt was $9.2 trillion, or 9.2 * 1012 dollars. At
that time, the U.S. population was approximately 303,000,000 (303 million), or
3.03 * 108. If the national debt was evenly divided among every individual in the
United States, how much would each citizen have to pay?
Solution The amount each citizen must pay is the total debt, 9.2 * 1012 dollars,
divided by the number of citizens, 3.03 * 108.
9.2 * 1012
9.2
1012
=
a
b
*
¢
≤
3.03
3.03 * 108
108
L 3.04 * 1012 - 8
= 3.04 * 104
= 30,400
Every U.S. citizen would have to pay approximately $30,400 to the federal government to pay off the national debt.
Check Point 6 Pell Grants help low-income undergraduate students pay for
college. In 2006, the federal cost of this program was $13 billion 1$13 * 1092 and
there were 5.1 million 15.1 * 1062 grant recipients. How much, to the nearest
hundred dollars, was the average grant?
An Application: Black Holes in Space
The concept of a black hole, a region in space where matter appears to vanish, intrigues
scientists and nonscientists alike. Scientists theorize that when massive stars run out of
nuclear fuel, they begin to collapse under the force of their own gravity. As the star
collapses, its density increases. In turn, the force of gravity increases so tremendously
that even light cannot escape from the star. Consequently, it appears black.
Section P.2 Exponents and Scientific Notation 27
A mathematical model, called the Schwarzchild formula, describes the critical
value to which the radius of a massive body must be reduced for it to become a
black hole. This model forms the basis of our next example.
EXAMPLE 7
An Application of Scientific Notation
Use the Schwarzchild formula
Rs =
where
2GM
c2
Rs = Radius of the star, in meters, that would cause it to become a black hole
M = Mass of the star, in kilograms
G = A constant, called the gravitational constant
m3
= 6.7 * 10-11
kg # s2
c = Speed of light
= 3 * 108 meters per second
to determine to what length the radius of the sun must be reduced for it to become
a black hole. The sun’s mass is approximately 2 * 1030 kilograms.
Solution
Rs =
=
=
=
=
=
2GM
c2
Use the given model.
2 * 6.7 * 10-11 * 2 * 1030
13 * 1082
2
G = 6.7 * 10-11, M = 2 * 1030,
and c = 3 * 108.
12 * 6.7 * 22 * 110-11 * 10302
13 * 1082
2
26.8 * 10-11 + 30
32 * 11082
2
26.8 * 1019
9 * 1016
Rearrange factors in the numerator.
Add exponents in the numerator. Raise each
factor in the denominator to the power.
Multiply powers to powers:
11082 = 108 2 = 1016.
2
26.8
* 1019 - 16
9
L 2.978 * 103
Substitute the given values:
#
When dividing expressions with the same
base, subtract the exponents.
Simplify.
= 2978
Although the sun is not massive enough to become a black hole (its radius is
approximately 700,000 kilometers), the Schwarzchild model theoretically indicates
that if the sun’s radius were reduced to approximately 2978 meters, that is, about
1
235,000 its present size, it would become a black hole.
7 The speed of blood, S, in centimeters per second, located
r centimeters from the central axis of an artery is modeled by
Check Point
S = 11.76 * 1052311.44 * 10-22 - r24.
Find the speed of blood at the central axis of this artery.
28 Chapter P Prerequisites: Fundamental Concepts of Algebra
Exercise Set P.2
Practice Exercises
Evaluate each exponential expression in Exercises 1–22.
1. 52 # 2
2. 62 # 2
3. 1 - 22
4. 1 -22
6
5. -2
4
6
6. -2 4
7. 1 - 320
8. 1 - 920
9. -30
5x3
≤
y
61. ¢
-15a4 b2
≤
5a10 b-3
63. ¢
3a -5 b2
≤
12a3 b-4
3
0
-3
60. ¢
3x4
≤
y
62. ¢
- 30a14 b8
≤
10a17 b-2
64. ¢
4a -5 b3
3 -5
12a b
≤
3
0
10. - 90
11. 4 -3
12. 2 -6
13. 2 2 # 2 3
14. 33 # 32
15. 12 22
16. 1332
3
In Exercises 65–76, write each number in decimal notation without
the use of exponents.
2
8
17.
-2
59. ¢
8
2
24
18.
3
34
19. 3 -3 # 3
20. 2 -3 # 2
23
21. 7
2
34
22. 7
3
Simplify each exponential expression in Exercises 23–64.
-2
23. x y
24. xy
-3
7 0
0 5
65. 3.8 * 102
66. 9.2 * 102
67. 6 * 10-4
68. 7 * 10-5
69. -7.16 * 106
70. - 8.17 * 106
71. 7.9 * 10-1
72. 6.8 * 10-1
73. -4.15 * 10-3
74. -3.14 * 10-3
75. -6.00001 * 1010
76. - 7.00001 * 1010
In Exercises 77–86, write each number in scientific notation.
77. 32,000
78. 64,000
25. x y
26. x y
79. 638,000,000,000,000,000
80. 579,000,000,000,000,000
27. x3 # x7
28. x11 # x5
81. - 5716
82. - 3829
29. x -5 # x10
30. x -6 # x12
83. 0.0027
84. 0.0083
31. 1x 2
32. 1x 2
85. -0.00000000504
86. -0.00000000405
x14
35. 7
x
x30
36. 10
x
x14
37. -7
x
x30
38. -10
x
3 7
11 5
33. 1x 2
34. 1x 2
-5 3
-6 4
39. 18x 2
4 2
3
6
42. a - b
y
3
43. 1 - 3x2y52
44. 1 -3x4 y62
47. 1- 9x3y21 - 2x6y42
48. 1 -5x4y21 -6x7 y112
2
45. 13x4212x 72
49.
8x20
2x4
46. 111x5219x122
50.
25a b
- 5a2b3
55. 14x32
57.
10x6
52.
35a b
- 7a7b3
54.
-2
24x3 y5
7 -9
32x y
92. 15.1 * 10-8213 * 10-42
93. 14.3 * 108216.2 * 1042
20b
10b20
56. 110x22
58.
95.
97.
99.
101.
10
14b
7b14
91. 16.1 * 10 -8212 * 10-42
20x24
14 6
7
53.
88. 12 * 104214.1 * 1032
3
13 4
51.
87. 13 * 104212.1 * 1032
89. 11.6 * 1015214 * 10-112
40. 16x 2
3 2
4
41. a - b
x
In Exercises 87–106, perform the indicated computations. Write the
answers in scientific notation. If necessary, round the decimal
factor in your scientific notation answer to two decimal places.
-3
10x 4 y9
30x12y -3
8.4 * 108
4 * 105
3.6 * 104
9 * 10-2
4.8 * 10-2
2.4 * 10
6
2.4 * 10-2
4.8 * 10
-6
90. 11.4 * 1015213 * 10-112
94. 18.2 * 108214.6 * 1042
96.
98.
100.
102.
6.9 * 108
3 * 105
1.2 * 104
2 * 10-2
7.5 * 10-2
2.5 * 106
1.5 * 10 -2
3 * 10-6
103.
480,000,000,000
0.00012
104.
282,000,000,000
0.00141
105.
0.00072 * 0.003
0.00024
106.
66,000 * 0.001
0.003 * 0.002
Section P.2 Exponents and Scientific Notation 29
Practice Plus
In Exercises 107–114, simplify each exponential expression.
Assume that variables represent nonzero real numbers.
-2
-3
1xy -22
1x -2 y2
107.
108.
3
-3
1x2y -12
1x -2 y2
x -3 y -4 z-5
≤
-2
112. ¢
-4 3 -2
12x -3 y -52
≤
Tickets Sold
1600
-3 3 0
2
12 -1 x -3 y -12 12x -6 y42 19x3y -32
-2
114.
x -4 y -5 z-6
United States Film Admissions
and Admission Charges
-4
12 x y 2 12x y 2 116x y 2
-1 -2 -1 -2
113.
x 4 y 5 z6
-2
0
12x y 2
-4 -6 2
Application Exercises
The bar graph shows the total amount Americans paid in federal
taxes, in trillions of dollars, and the U.S. population, in millions,
from 2003 through 2006. Exercises 115–116 are based on the
numbers displayed by the graph.
Annual Admissions
(billions of tickets sold)
111. ¢
x 3 y 4 z5
110. 13x -4 yz-7213x2-3
1400
1200
1000
Price per Ticket
1420
1400
1260
1190
$8
1450
6.40
6.60
4.20
$7
$6
5.40
$5
4.40
800
$4
600
$3
400
$2
200
$1
1990
1995
2000
Year
2005
Average Price per Ticket
109. 12x -3 yz-6212x2-5
In the dramatic arts, ours is the era of the movies. As individuals
and as a nation, we’ve grown up with them. Our images of love,
war, family, country—even of things that terrify us—owe much to
what we’ve seen on screen. The bar graph quantifies our love for
movies by showing the number of tickets sold, in billions, and the
average price per ticket for five selected years. Exercises 117–118
are based on the numbers displayed by the graph.
2006
Source: National Association of Theater Owners
Federal Taxes and the United States Population
Federal Taxes Collected
Population
Federal Taxes Collected
(trillions of dollars)
$2.00
295
292
$2.50
1.95
2.27
298
2.52 300
2.02
300
250
$1.50
200
$1.00
150
$0.50
100
2003
2005
2004
Population (millions)
350
$3.00
2006
Year
Sources: Internal Revenue Service and U.S. Census Bureau
115. a. In 2006, the United States government collected
$2.52 trillion in taxes. Express this number in scientific
notation.
b. In 2006, the population of the United States was
approximately 300 million. Express this number in
scientific notation.
c. Use your scientific notation answers from parts (a) and
(b) to answer this question: If the total 2006 tax
collections were evenly divided among all Americans,
how much would each citizen pay? Express the answer
in decimal notation.
116. a. In 2005, the United States government collected
$2.27 trillion in taxes. Express this number in scientific
notation.
b. In 2005, the population of the United States was
approximately 298 million. Express this number in
scientific notation.
c. Use your scientific notation answers from parts (a) and
(b) to answer this question: If the total 2005 tax
collections were evenly divided among all Americans,
how much would each citizen pay? Express the answer
in decimal notation, rounded to the nearest dollar.
117. Use scientific notation to compute the amount of money
that the motion picture industry made from box-office
receipts in 2006. Express the answer in scientific notation.
118. Use scientific notation to compute the amount of money
that the motion picture industry made from box office
receipts in 2005. Express the answer in scientific notation.
119. The mass of one oxygen molecule is 5.3 * 10-23 gram. Find
the mass of 20,000 molecules of oxygen. Express the answer
in scientific notation.
120. The mass of one hydrogen atom is 1.67 * 10-24 gram. Find
the mass of 80,000 hydrogen atoms. Express the answer in
scientific notation.
121. In this exercise, use the fact that there are approximately
3.2 * 107 seconds in a year. According to the United States
Department of Agriculture, Americans consume 127
chickens per second. How many chickens are eaten per year
in the United States? Express the answer in scientific
notation.
122. Convert 365 days (one year) to hours, to minutes, and,
finally, to seconds, to determine how many seconds there
are in a year. Express the answer in scientific notation.
Writing in Mathematics
123. Describe what it means to raise a number to a power. In
your description, include a discussion of the difference
between -52 and 1- 522.
124. Explain the product rule for exponents. Use 2 3 # 2 5 in your
explanation.
4
125. Explain the power rule for exponents. Use 1322 in your
explanation.
58
126. Explain the quotient rule for exponents. Use 2 in your
5
explanation.
127. Why is 1-3x2212x -52 not simplified? What must be done to
simplify the expression?
128. How do you know if a number is written in scientific
notation?
30 Chapter P Prerequisites: Fundamental Concepts of Algebra
129. Explain how to convert from scientific to decimal notation
and give an example.
130. Explain how to convert from decimal to scientific notation
and give an example.
Group Exercise
Critical Thinking Exercises
Make Sense? In Exercises 131–134, determine whether each
statement makes sense or does not make sense, and explain your
reasoning.
131. There are many exponential expressions that are equal to
2
9
6
36x12, such as 16x62 , 16x3216x92, 361x32 , and 621x22 .
-2
132. If 5 is raised to the third power, the result is a number
between 0 and 1.
133. The population of Colorado is approximately 4.6 * 1012.
134. I just finished reading a book that contained approximately
1.04 * 105 words.
In Exercises 135–142, determine whether each statement is true or
false. If the statement is false, make the necessary change(s) to
produce a true statement.
135. 4
-2
6 4
-3
137. 1-22 = 2
4
-4
139. 534.7 = 5.347 * 103
136. 5
-2
138. 5
2
140.
145. Our hearts beat approximately 70 times per minute.
Express in scientific notation how many times the heart
beats over a lifetime of 80 years. Round the decimal factor
in your scientific notation answer to two decimal places.
7 2
#5
-2
8 * 10
-5
7 2
30
4 * 10-5
5
#2
-5
= 2 * 1025
141. 17 * 1052 + 12 * 10-32 = 9 * 102
146. Putting Numbers into Perspective. A large number can be
put into perspective by comparing it with another number.
For example, we put the $9.2 trillion national debt
(Example 6) and the $2.52 trillion the government collected
in taxes (Exercise 115) by comparing these numbers to the
number of U.S. citizens.
For this project, each group member should consult an
almanac, a newspaper, or the World Wide Web to find a
number greater than one million. Explain to other members
of the group the context in which the large number is used.
Express the number in scientific notation. Then put the
number into perspective by comparing it with another
number.
Preview Exercises
Exercises 147–149 will help you prepare for the material covered
in the next section.
147. a. Find 216 # 24.
b. Find 216 # 4 .
c. Based on your answers to parts (a) and (b), what can
you conclude?
143. The mad Dr. Frankenstein has gathered enough bits and
pieces (so to speak) for 2 -1 + 2 -2 of his creature-to-be.
Write a fraction that represents the amount of his creature
that must still be obtained.
148. a. Use a calculator to approximate 2300 to two decimal
places.
b. Use a calculator to approximate 10 23 to two decimal
places.
c. Based on your answers to parts (a) and (b), what can
you conclude?
144. If bA = MN, bC = M, and bD = N, what is the relationship among A, C, and D?
149. a. Simplify:
b. Simplify:
142. 14 * 1032 + 13 * 1022 = 4.3 * 103
Section
P.3
Objectives
� Evaluate square roots.
� Simplify expressions of the
form 2a 2 .
� Use the product rule to
simplify square roots.
� Use the quotient rule to
simplify square roots.
� Add and subtract square
roots.
� Rationalize denominators.
� Evaluate and perform
operations with higher roots.
� Understand and use rational
exponents.
21x + 10x.
2122 + 1022.
Radicals and Rational Exponents
T
his photograph shows mathematical models
used by Albert Einstein at a lecture on
relativity. Notice the radicals that appear in
many of the formulas. Among these
models, there is one describing
how an astronaut in a moving
spaceship ages more slowly than
friends who remain on Earth.
No description of your world
can be complete without roots
and radicals. In this section, in
addition to reviewing the
basics of radical expressions
and the use of rational
exponents to indicate radicals,
you will see how radicals model
time dilation for a futuristic highspeed trip to a nearby star.
46 Chapter P Prerequisites: Fundamental Concepts of Algebra
Section
P.4
Objectives
� Understand the vocabulary of
polynomials.
� Add and subtract
polynomials.
� Multiply polynomials.
� Use FOIL in polynomial
multiplication.
� Use special products in
polynomial multiplication.
� Perform operations with
polynomials in several
variables.
Polynomials
C
an that be Axl, your author’s yellow lab, sharing a special
moment with a baby chick? And if it is (it is), what possible
relevance can this have to polynomials? An answer is
promised before you reach the exercise set. For now, we
open the section by defining and describing
polynomials.
How We Define Polynomials
More education results in a higher income.
The mathematical models
Old Dog Á New Chicks
M = - 18x3 + 923x2 - 9603x + 48,446
and฀ W = 17x3 - 450x2 + 6392x - 14,764
describe the median, or middlemost, annual income for men, M, and women, W,
who have completed x years of education. We’ll be working with these models and
the data upon which they are based in the exercise set.
The algebraic expressions that appear on the right sides of the models are
examples of polynomials. A polynomial is a single term or the sum of two or more
terms containing variables with whole-number exponents. The polynomials above
each contain four terms. Equations containing polynomials are used in such diverse
areas as science, business, medicine, psychology, and sociology. In this section, we
review basic ideas about polynomials and their operations.
�
Understand the vocabulary of
polynomials.
How We Describe Polynomials
Consider the polynomial
7x3 - 9x2 + 13x - 6.
We can express this polynomial as
7x3 + 1- 9x22 + 13x + 1 - 62.
The polynomial contains four terms. It is customary to write the terms in the
order of descending powers of the variable. This is the standard form of a
polynomial.
Some polynomials contain only one variable. Each term of such a polynomial
in x is of the form axn. If a Z 0, the degree of ax n is n. For example, the degree of
the term 7x3 is 3.
Study Tip
We can express 0 in many ways,
including 0x, 0x2, and 0x3. It is
impossible to assign a unique
exponent to the variable. This is why
0 has no defined degree.
The Degree of axn
If a Z 0, the degree of axn is n. The degree of a nonzero constant is 0. The
constant 0 has no defined degree.
Section P.4 Polynomials
47
Here is an example of a polynomial and the degree of each of its four terms:
6x4-3x3+2x-5.
degree 4
degree 3
degree 1
degree of nonzero constant: 0
Notice that the exponent on x for the term 2x, meaning 2x1, is understood to be 1.
For this reason, the degree of 2x is 1. You can think of - 5 as - 5x0; thus, its degree
is 0.
A polynomial is simplified when it contains no grouping symbols and no like
terms. A simplified polynomial that has exactly one term is called a monomial. A
binomial is a simplified polynomial that has two terms. A trinomial is a simplified
polynomial with three terms. Simplified polynomials with four or more terms have
no special names.
The degree of a polynomial is the greatest degree of all the terms of the
polynomial. For example, 4x2 + 3x is a binomial of degree 2 because the degree of
the first term is 2, and the degree of the other term is less than 2. Also,
7x5 - 2x2 + 4 is a trinomial of degree 5 because the degree of the first term is 5,
and the degrees of the other terms are less than 5.
Up to now, we have used x to represent the variable in a polynomial. However,
any letter can be used. For example,
• 7x5 - 3x3 + 8
is a polynomial (in x) of degree 5. Because there are
three terms, the polynomial is a trinomial.
• 6y3 + 4y2 - y + 3
is a polynomial (in y) of degree 3. Because there are
four terms, the polynomial has no special name.
• z7 + 22
is a polynomial (in z) of degree 7. Because there are
two terms, the polynomial is a binomial.
We can tie together the threads of our discussion with the formal definition of
a polynomial in one variable. In this definition, the coefficients of the terms are
represented by a n (read “a sub n”), an - 1 (read “a sub n minus 1”), an - 2 , and so on.
The small letters to the lower right of each a are called subscripts and are not
exponents. Subscripts are used to distinguish one constant from another when a
large and undetermined number of such constants are needed.
Definition of a Polynomial in x
A polynomial in x is an algebraic expression of the form
anxn + an - 1xn - 1 + an - 2xn - 2 + Á + a1x + a0 ,
where an , an - 1 , an - 2 , Á , a1 , and a0 are real numbers, an Z 0, and n is a nonnegative integer. The polynomial is of degree n, an is the leading coefficient, and
a 0 is the constant term.
�
Add and subtract polynomials.
Adding and Subtracting Polynomials
Polynomials are added and subtracted by combining like terms. For example, we can
combine the monomials -9x3 and 13x3 using addition as follows:
–9x3+13x3=(–9+13)x3=4x3.
These like terms both
contain x to the
third power.
Add coefficients
and keep the same
variable factor, x3.
48 Chapter P Prerequisites: Fundamental Concepts of Algebra
Adding and Subtracting Polynomials
EXAMPLE 1
Perform the indicated operations and simplify:
a. 1 -9x3 + 7x2 - 5x + 32 + 113x3 + 2x2 - 8x - 62
b. 17x3 - 8x2 + 9x - 62 - 12x3 - 6x2 - 3x + 92.
Solution
a. 1-9x3 + 7x2 - 5x + 32 + 113x3 + 2x2 - 8x - 62
= 1 -9x3 + 13x32 + 17x2 + 2x22 + 1-5x - 8x2 + 13 - 62 Group like terms.
= 4x3 + 9x2 + 1-13x2 + 1 - 32
3
2
3
2
Combine like terms.
= 4x + 9x - 13x - 3
Study Tip
b.
2
7x - 8x + 9x - 6
-2x3 + 6x2 + 3x - 9
5x3 - 2x2 + 12x - 15
The like terms can be combined by
adding their coefficients and keeping
the same variable factor.
2
(7x -8x +9x-6)-(2x -6x -3x+9)
You can also arrange like terms in
columns and combine vertically:
3
Simplify.
3
Change the sign of each coefficient.
=(7x3-8x2+9x-6)+(–2x3+6x2+3x-9)
Rewrite subtraction
as addition of the
additive inverse.
= 17x3 - 2x32 + 1-8x2 + 6x22 + 19x + 3x2 + 1- 6 - 92 Group like terms.
= 5x3 + 1-2x22 + 12x + 1-152
3
Combine like terms.
2
= 5x - 2x + 12x - 15
Check Point
1
Simplify.
Perform the indicated operations and simplify:
a. 1 -17x3 + 4x2 - 11x - 52 + 116x3 - 3x2 + 3x - 152
b. 113x3 - 9x2 - 7x + 12 - 1 -7x3 + 2x2 - 5x + 92.
�
Multiply polynomials.
Multiplying Polynomials
The product of two monomials is obtained by using properties of exponents. For
example,
Study Tip
Don’t confuse adding and multiplying
monomials.
(–8x6)(5x3)=–8 ⴢ 5x6±3=–40x9.
Addition:
Multiply coefficients and add exponents.
5x4 + 6x4 = 11x4
Multiplication:
15x4216x42 = 15 # 621x4 # x42
Furthermore, we can use the distributive property to multiply a monomial and a
polynomial that is not a monomial. For example,
= 30x4 + 4
= 30x8
Only like terms can be added or
subtracted, but unlike terms may be
multiplied.
3x4(2x3-7x+3)=3x4 ⴢ 2x3-3x4 ⴢ 7x+3x4 ⴢ 3=6x7-21x5+9x4.
Monomial
Trinomial
Addition:
5x4 + 3x2 cannot be simplified.
Multiplication:
15x4213x22 = 15 # 321x4 # x22
How do we multiply two polynomials if neither is a monomial? For example,
consider
(2x+3)(x2+4x +5).
= 15x4 + 2
= 15x6
Binomial
Trinomial
Section P.4 Polynomials
49
One way to perform (2x + 3)(x2 + 4x + 5) is to distribute 2x throughout the
trinomial
2x1x2 + 4x + 52
and 3 throughout the trinomial
31x2 + 4x + 52.
Then combine the like terms that result.
Multiplying Polynomials When Neither Is a Monomial
Multiply each term of one polynomial by each term of the other polynomial.
Then combine like terms.
Multiplying a Binomial and a Trinomial
EXAMPLE 2
Multiply:
Solution
12x + 321x2 + 4x + 52.
12x + 321x2 + 4x + 52
= 2x1x2 + 4x + 52 + 31x2 + 4x + 52
= 2x # x2 + 2x # 4x + 2x # 5 + 3x2 + 3 # 4x + 3 # 5
3
2
2
Multiply the trinomial by each
term of the binomial.
Use the distributive property.
= 2x + 8x + 10x + 3x + 12x + 15
Multiply monomials: Multiply
coefficients and add
exponents.
= 2x3 + 11x2 + 22x + 15
Combine like terms:
8x2 + 3x2 = 11x2 and
10x + 12x = 22x.
Another method for performing the multiplication is to use a vertical format
similar to that used for multiplying whole numbers.
x2+4x+ 5
2x+ 3
3x2+12x+15
Write like terms in the
same column.
2x3+ 8x2+10x
Combine like terms.
Check Point
�
Use FOIL in polynomial
multiplication.
2
3
3(x2 + 4x + 5)
2x(x2 + 4x + 5)
2
2x +11x +22x+15
Multiply:
15x - 2213x2 - 5x + 42.
The Product of Two Binomials: FOIL
Frequently, we need to find the product of two binomials. One way to perform this
multiplication is to distribute each term in the first binomial through the second
binomial. For example, we can find the product of the binomials 3x + 2 and 4x + 5
as follows:
(3x+2)(4x+5)=3x(4x+5)+2(4x+5)
=3x(4x)+3x(5)+2(4x)+2(5)
Distribute 3x
Distribute 2
=12x2+15x+8x+10.
over 4x + 5.
over 4x + 5.
We'll combine these like terms later.
For now, our interest is in how to obtain
each of these four terms.
50 Chapter P Prerequisites: Fundamental Concepts of Algebra
We can also find the product of 3x + 2 and 4x + 5 using a method called
FOIL, which is based on our work shown at the bottom of the previous page. Any
two binomials can be quickly multiplied by using the FOIL method, in which F
represents the product of the first terms in each binomial, O represents the product of the outside terms, I represents the product of the inside terms, and L represents the product of the last, or second, terms in each binomial. For example, we
can use the FOIL method to find the product of the binomials 3x + 2 and 4x + 5
as follows:
last
first
F
O
I
L
(3x+2)(4x+5)=12x2+15x+8x+10
inside
outside
Product
of
First
terms
Product
of
Outside
terms
Product
of
Inside
terms
Product
of
Last
terms
=12x2+23x+10
Combine like terms.
In general, here’s how to use the FOIL method to find the product of ax + b
and cx + d:
Using the FOIL Method to Multiply Binomials
last
first
F
O
I
L
(ax+b)(cx+d)=ax ⴢ cx+ax ⴢ d+b ⴢ cx+b ⴢ d
inside
outside
Solution
Product
of
Outside
terms
Product
of
Inside
terms
Product
of
Last
terms
Using the FOIL Method
EXAMPLE 3
Multiply:
Product
of
First
terms
13x + 4215x - 32.
last
first
F
O
I
L
(3x+4)(5x-3)=3x ⴢ 5x+3x(–3)+4 ⴢ 5x+4(–3)
inside
outside
Check Point
�
Use special products in
polynomial multiplication.
3
Multiply:
=15x2-9x+20x-12
=15x2+11x-12
Combine like terms.
17x - 5214x - 32.
Special Products
There are several products that occur so frequently that it’s convenient to memorize
the form, or pattern, of these formulas.
Section P.4 Polynomials
51
Special Products
Let A and B represent real numbers, variables, or algebraic expressions.
Study Tip
Although it’s convenient to memorize
these forms, the FOIL method can be
used on all five examples in the box.
To cube x + 4, you can first square
x + 4 using FOIL and then multiply
this result by x + 4. In short, you do
not necessarily have to utilize these
special formulas. What is the advantage of knowing and using these
forms?
Special Product
Example
Sum and Difference of Two Terms
1A + B21A - B2 = A2 - B2
12x + 3212x - 32 = 12x22 - 32
= 4x2 - 9
Squaring a Binomial
1A + B22 = A2 + 2AB + B2
1y + 522 = y2 + 2 # y # 5 + 52
1A - B22 = A2 - 2AB + B2
13x - 422
= y2 + 10y + 25
= 13x22 - 2 # 3x # 4 + 4 2
= 9x2 - 24x + 16
Cubing a Binomial
1A + B23 = A3 + 3A2B + 3AB2 + B3
1x + 423
= x3 + 3x2142 + 3x1422 + 4 3
= x3 + 12x2 + 48x + 64
1A - B23 = A3 - 3A2B + 3AB2 - B3
1x - 223
= x3 - 3x2122 + 3x1222 - 2 3
= x3 - 6x2 + 12x - 8
�
Perform operations with
polynomials in several variables.
Polynomials in Several Variables
A polynomial in two variables, x and y, contains the sum of one or more monomials
in the form axnym. The constant, a, is the coefficient. The exponents, n and m,
represent whole numbers. The degree of the monomial axnym is n + m.
Here is an example of a polynomial in two variables:
The coefficients are 7, −17, 1, −6, and 9.
7x2y3
Degree of
monomial:
2+3=5
-
17x4y2
Degree of
monomial:
4+2=6
+
xy
Degree of
monomial A1x1y1B:
1+1=2
-
6y2
+
Degree of
monomial A−6x0y2B:
0+2=2
9.
Degree of
monomial A9x0y0B:
0+0=0
The degree of a polynomial in two variables is the highest degree of all its
terms. For the preceding polynomial, the degree is 6.
Polynomials containing two or more variables can be added, subtracted, and
multiplied just like polynomials that contain only one variable. For example, we can
add the monomials -7xy2 and 13xy2 as follows:
–7xy2+13xy2=(–7+13)xy2=6xy2.
These like terms both contain
the variable factors x and y2.
Add coefficients and keep the
same variable factors, xy2.
52 Chapter P Prerequisites: Fundamental Concepts of Algebra
Multiplying Polynomials in Two Variables
EXAMPLE 4
b. 15x + 3y22.
Multiply: a. 1x + 4y213x - 5y2
Solution We will perform the multiplication in part (a) using the FOIL method.
We will multiply in part (b) using the formula for the square of a binomial sum,
1A + B22.
a. (x+4y)(3x-5y)
F
Multiply these binomials using the FOIL method.
O
I
L
=(x)(3x)+(x)(–5y)+(4y)(3x)+(4y)(–5y)
=3x2-5xy+12xy-20y2
=3x2+7xy-20y2 Combine like terms.
=
(A + B)2
A2
+
2 ⴢ A ⴢ B + B2
b. (5x+3y)2=(5x)2+2(5x)(3y)+(3y)2
=25x2+30xy+9y2
Check Point
4
Multiply:
a. 17x - 6y213x - y2
b. 12x + 4y22.
Special products can sometimes be used to find the products of certain
trinomials, as illustrated in Example 5.
Using the Special Products
EXAMPLE 5
Multiply:
a. 17x + 5 + 4y217x + 5 - 4y2
b. 13x + y + 122.
Solution
a. By grouping the first two terms within each of the parentheses, we can find the
product using the form for the sum and difference of two terms.
+
(A
B)
(A
ⴢ
−
=
B)
A2
−
B2
[(7x+5)+4y] ⴢ [(7x+5)-4y]=(7x+5)2-(4y)2
= 17x22 + 2 # 7x # 5 + 52 - 14y22
= 49x2 + 70x + 25 - 16y2
b. We can group the terms of 13x + y + 12 so that the formula for the square of
a binomial can be applied.
2
(A
+
B)2
=
A2
+
2
ⴢ
A
ⴢ
B
+
B2
[(3x+y)+1]2 =(3x+y)2+2 ⴢ (3x+y) ⴢ 1+12
= 9x2 + 6xy + y2 + 6x + 2y + 1
Check Point
5
Multiply:
a. 13x + 2 + 5y213x + 2 - 5y2
b. 12x + y + 322.
Section P.4 Polynomials
53
Labrador Retrievers and Polynomial Multiplication
The color of a Labrador retriever is determined by its pair of genes. A single gene is inherited at
random from each parent. The black-fur gene, B, is dominant. The yellow-fur gene, Y, is recessive.
This means that labs with at least one black-fur gene (BB or BY) have black coats. Only labs with
two yellow-fur genes (YY) have yellow coats.
Axl, your author’s yellow lab, inherited his genetic makeup from two black BY parents.
Second BY parent, a black lab
with a recessive yellow-fur gene
First BY parent, a black lab
with a recessive yellow-fur gene
B
Y
B
BB
BY
Y
BY YY
The table shows the four possible
combinations of color genes that BY
parents can pass to their offspring.
Because YY is one of four possible outcomes, the probability that a yellow lab like Axl will be the offspring of these black parents
is 14 .
The probabilities suggested by the table can be modeled by the expression A 12 B + 12 Y B 2.
2
2
1
1
1
1
1
1
a B+ Yb = a Bb +2 a Bb a Yb + a Yb
2
2
2
2
2
2
1
BB +
4
=
The probability of a
black lab with two
dominant black genes is 1 .
4
1
BY
2
2
1
+ YY
4
The probability of a
black lab with a
recessive yellow gene is 1 .
2
The probability of a
yellow lab with two
recessive yellow genes is 1 .
4
Exercise Set P.4
Practice Exercises
In Exercises 15–82, find each product.
In Exercises 1–4, is the algebraic expression a polynomial? If it is,
write the polynomial in standard form.
2
2. 2x + 3x
1. 2x + 3x - 5
2x + 3
3.
x
2
-1
- 5
3
4
4. x - x + x - 5
In Exercises 5–8, find the degree of the polynomial.
2
3
6. -4x + 7x - 11
5. 3x - 5x + 4
2
3
2
4
7. x - 4x + 9x - 12x + 63
8. x2 - 8x3 + 15x4 + 91
In Exercises 9–14, perform the indicated operations. Write the
resulting polynomial in standard form and indicate its degree.
9. 1-6x3 + 5x2 - 8x + 92 + 117x3 + 2x2 - 4x - 132
10. 1-7x3 + 6x2 - 11x + 132 + 119x3 - 11x2 + 7x - 172
11. 117x3 - 5x2 + 4x - 32 - 15x3 - 9x2 - 8x + 112
12. 118x4 - 2x3 - 7x + 82 - 19x4 - 6x3 - 5x + 72
13. 15x2 - 7x - 82 + 12x2 - 3x + 72 - 1x2 - 4x - 32
14. 18x2 + 7x - 52 - 13x2 - 4x2 - 1- 6x3 - 5x2 + 32
15. 1x + 121x2 - x + 12
16. 1x + 521x2 - 5x + 252
19. 1x + 721x + 32
20. 1x + 821x + 52
17. 12x - 321x2 - 3x + 52
21. 1x - 521x + 32
23. 13x + 5212x + 12
25. 12x - 3215x + 32
27. 15x - 4213x - 72
2
2
29. 18x3 + 321x2 - 52
31. 1x + 321x - 32
33. 13x + 2213x - 22
35. 15 - 7x215 + 7x2
37. 14x2 + 5x214x2 - 5x2
39. 11 - y5211 + y52
41. 1x + 222
18. 12x - 121x2 - 4x + 32
22. 1x - 121x + 22
24. 17x + 4213x + 12
26. 12x - 5217x + 22
28. 17x2 - 2213x2 - 52
30. 17x3 + 521x2 - 22
32. 1x + 521x - 52
34. 12x + 5212x - 52
36. 14 - 3x214 + 3x2
38. 13x2 + 4x213x2 - 4x2
40. 12 - y5212 + y52
42. 1x + 522
54 Chapter P Prerequisites: Fundamental Concepts of Algebra
44. 13x + 222
46. 1x - 422
48. 15x2 - 32
2
50. 19 - 5x22
Men
52. 1x + 22
56. 1x - 123
57. 13x - 42
3
58. 12x - 32
3
59. 1x + 5y217x + 3y2
61. 1x - 3y212x + 7y2
63. 13xy - 1215xy + 22
65. 17x + 5y22
67. 1x2y2 - 32
2
69. 1x - y21x2 + xy + y22
71. 13x + 5y213x - 5y2
73. 1x + y + 321x + y - 32
60. 1x + 9y216x + 7y2
62. 13x - y212x + 5y2
64. 17x2 y + 1212x2 y - 32
66. 19x + 7y22
68. 1x2 y 2 - 52
2
70. 1x + y21x2 - xy + y22
72. 17x + 3y217x - 3y2
75. 13x + 7 - 5y213x + 7 + 5y2
78. 38y + 17 - 3x2438y - 17 - 3x24
79. 1x + y + 122
80. 1x + y + 222
81. 12x + y + 122
82. 15x + 1 + 6y22
Practice Plus
In Exercises 83–90, perform the indicated operation or operations.
83. 13x + 4y22 - 13x - 4y22
84. 15x + 2y22 - 15x - 2y22
85. 15x - 7213x - 22 - 14x - 5216x - 12
86. 13x + 5212x - 92 - 17x - 221x - 12
87. 12x + 5212x - 5214x2 + 252
88. 13x + 4213x - 4219x2 + 162
12x - 72
$40
$30
$20
$10
8
10
12
13
14
16
Years of School Completed
18
20
Source: Bureau of the Census
Here are polynomial models that describe the median annual
income for men, M, and for women, W, who have completed x
years of education:
M = 177x2 + 288x + 7075
W = 255x2 - 2956x + 24,336
91. a. Use the equation defined by a polynomial of degree 2 to
find the median annual income for a man with 16 years of
education. Does this underestimate or overestimate the
median income shown by the bar graph? By how much?
b. Use the equations defined by polynomials of degree 3 to
find a mathematical model for M - W.
c. According to the model in part (b), what is the difference
in the median annual income between men and women
with 14 years of education?
d. According to the data displayed by the graph, what is the
actual difference in the median annual income between
men and women with 14 years of education? Did the
result of part (c) underestimate or overestimate this
difference? By how much?
77. 35y - 12x + 32435y + 12x + 324
3
$50
Exercises 91–92 are based on these models and the data displayed
by the graph.
76. 15x + 7y - 2215x + 7y + 22
12x - 725
$60
M = - 18x3 + 923x2 - 9603x + 48,446
W = 17x3 - 450x2 + 6392x - 14,764.
74. 1x + y + 521x + y - 52
89.
$70
82,401
68,875
55. 1x - 323
$80
71,530
54. 13x + 423
51,316
53. 12x + 323
Women
$90
3
Median Annual Income
(thousands of dollars)
3
57,220
41,681
51. 1x + 12
Median Annual Income, by Level of Education, 2004
44,404
33,481
49. 17 - 2x22
41,895
30,816
2
35,725
26,029
47. 14x2 - 12
As you complete more years of education, you can count on a
greater income. The bar graph shows the median, or middlemost,
annual income for Americans, by level of education, in 2004.
26,277
19,169
45. 1x - 322
Application Exercises
21,659
17,023
43. 12x + 322
90.
15x - 326
15x - 324
92. a. Use the equation defined by a polynomial of degree 2 to
find the median annual income for a woman with
18 years of education. Does this underestimate or overestimate the median income shown by the bar graph? By
how much?
b. Use the equations defined by polynomials of degree 3 to
find a mathematical model for M - W.
c. According to the model in part (b), what is the difference
in the median annual income between men and women
with 16 years of education?
d. According to the data displayed by the graph, what is the
actual difference in the median annual income between
men and women with 16 years of education? Did the
result of part (c) underestimate or overestimate this
difference? By how much?
Section P.4 Polynomials
The volume, V, of a rectangular solid with length l, width w, and
height h is given by the formula V = lwh. In Exercises 93–94, use
this formula to write a polynomial in standard form that models,
or represents, the volume of the open box.
93.
x
8 − 2x
10 − 2x
55
104. I used the FOIL method to find the product of x + 5 and
x2 + 2x + 1.
105. Many English words have prefixes with meanings similar to
those used to describe polynomials, such as monologue,
binocular, and tricuspid.
106. Special-product formulas have patterns that make their
multiplications quicker than using the FOIL method.
107. Express the area of the plane figure shown as a polynomial
in standard form.
94.
x
x
8 − 2x
x
x−1
5 − 2x
x+3
In Exercises 95–96, write a polynomial in standard form that
models, or represents, the area of the shaded region.
95.
x+9
x+3
x+5
x+1
In Exercises 108–109, represent the volume of each figure as a
polynomial in standard form.
108.
x
96.
x+4
x+3
x+1
2x − 1
x+2
x+1
x
x+3
109.
Writing in Mathematics
97. What is a polynomial in x?
98. Explain how to subtract polynomials.
99. Explain how to multiply two binomials using the FOIL
method. Give an example with your explanation.
100. Explain how to find the product of the sum and difference
of two terms. Give an example with your explanation.
101. Explain how to square a binomial difference. Give an
example with your explanation.
x+2
3
x
x+5
2x + 1
110. Simplify: 1yn + 221yn - 22 - 1yn - 32 .
2
102. Explain how to find the degree of a polynomial in two
variables.
Preview Exercises
Critical Thinking Exercises
Exercises 111–113 will help you prepare for the material covered
in the next section. In each exercise, replace the boxed question
mark with an integer that results in the given product. Some trial
and error may be necessary.
Make Sense? In Exercises 103–106, determine whether each
statement makes sense or does not make sense, and explain your
reasoning.
103. Knowing the difference between factors and terms is
2
important: In 13x2 y2 , I can distribute the exponent 2 on
2
each factor, but in 13x2 + y2 , I cannot do the same thing
on each term.
111. 1x + 321x + 冷฀ ?฀ 冷2 = x2 + 7x + 12
112. 1x - 冷฀ ?฀ 冷21x - 122 = x2 - 14x + 24
113. 14x + 1212x - 冷฀ ?฀ 冷2 = 8x2 - 10x - 3
82 Chapter P Prerequisites: Fundamental Concepts of Algebra
119. 1x - y2-1 + 1x - y2-2
111. When performing the division
1x + 32
7x
,
,
x + 3
x - 5
2
I began by dividing the numerator and the denominator by
the common factor, x + 3.
112. I subtracted
3x - 5
x - 3
from
and obtained a constant.
x - 1
x - 1
In Exercises 113–116, determine whether each statement is true or
false. If the statement is false, make the necessary change(s) to
produce a true statement.
x2 - 25
= x - 5
113.
x - 5
- 3y - 6
114. The expression
simplifies to the consecutive
y + 2
integer that follows - 4.
2x - 1
3x - 1
5x - 2
+
= 0
x - 7
x - 7
x - 7
1
7
=
116. 6 +
x
x
115.
1
1
1
- n
- 2n
xn - 1
x + 1
x - 1
1
1
1
1
b a1 b a1 b
118. a 1 - b a1 x
x + 1
x + 2
x + 3
Objectives
� Solve linear equations in
�
�
�
�
�
�
�
�
쐅
쐈
쐉
one variable.
Solve linear equations
containing fractions.
Solve rational equations with
variables in the denominators.
Solve a formula for a variable.
Solve equations involving
absolute value.
Solve quadratic equations by
factoring.
Solve quadratic equations by
the square root property.
Solve quadratic equations by
completing the square.
Solve quadratic equations
using the quadratic formula.
Use the discriminant to
determine the number and
type of solutions of quadratic
equations.
Determine the most efficient
method to use when solving a
quadratic equation.
Solve radical equations.
Preview Exercises
Exercises 121–123 will help you prepare for the material covered
in the next section.
121. If 6 is substituted for x in the equation
2(x - 3) - 17 = 13 - 3 (x + 2),
122. Multiply and simplify:
117.
P.7
does to each number x.
is the resulting statement true or false?
In Exercises 117–119, perform the indicated operations.
Section
120. In one short sentence, five words or less, explain what
1
1
1
+ 2 + 3
x
x
x
1
1
1
+ 5 + 6
x4
x
x
12a
x + 2
x - 1
b.
4
3
- b - 2b2 - 4ac
2a
for a = 2, b = 9, and c = - 5.
123. Evaluate
Equations
M
ath tattoos. Who knew?
Do you recognize the
significance of this tattoo?
The algebraic expression gives
the solutions of a quadratic
equation.
In this section, we will
review how to solve a variety of
equations, including linear
equations, quadratic equations,
and radical equations.
Solving Linear Equations in One Variable
We begin with a general definition of a linear equation in one variable.
Definition of a Linear Equation
A linear equation in one variable x is an equation that can be written in the form
ax + b = 0,
where a and b are real numbers, and a Z 0.
An example of a linear equation in one variable is
4x + 12 = 0.
Section P.7 Equations 83
Solving an equation in x involves determining all values of x that result in a true
statement when substituted into the equation. Such values are solutions, or roots, of
the equation. For example, substitute - 3 for x in 4x + 12 = 0. We obtain
41 - 32 + 12 = 0,฀ or฀ -12 + 12 = 0.
This simplifies to the true statement 0 = 0. Thus, -3 is a solution of the equation
4x + 12 = 0. We also say that -3 satisfies the equation 4x + 12 = 0, because when
we substitute - 3 for x, a true statement results. The set of all such solutions is called
the equation’s solution set. For example, the solution set of the equation
4x + 12 = 0 is 5 -36 because -3 is the equation’s only solution.
Two or more equations that have the same solution set are called equivalent
equations. For example, the equations
4x + 12 = 0฀ and฀ 4x = - 12฀ and฀ x = - 3
are equivalent equations because the solution set for each is 5 -36. To solve a linear
equation in x, we transform the equation into an equivalent equation one or more
times. Our final equivalent equation should be of the form
x = a number.
The solution set of this equation is the set consisting of the number.
To generate equivalent equations, we will use the following principles:
Generating Equivalent Equations
An equation can be transformed into an equivalent equation by one or more of
the following operations:
Example
1. Simplify an expression
by removing grouping
symbols and combining
like terms.
•
3(x-6)=6x-x
3x - 18 = 5x
3x - 18 = 5x
2. Add (or subtract) the
same real number or
variable expression on
both sides of the
equation.
•
3. Multiply (or divide) by
the same nonzero
quantity on both sides of
the equation.
•
–18=2x
2x
–18
=
2
2
-9 = x
4. Interchange the two
sides of the equation.
•
-9 = x
x = -9
3x-18-3x=5x-3x
Subtract 3x from
both sides of the
equation.
-18 = 2x
Divide both sides
of the equation
by 2.
If you look closely at the equations in the box, you will notice that we have
solved the equation 31x - 62 = 6x - x. The final equation, x = - 9, with x
isolated on the left side, shows that 5- 96 is the solution set. The idea in solving a
linear equation is to get the variable by itself on one side of the equal sign and a
number by itself on the other side.
84 Chapter P Prerequisites: Fundamental Concepts of Algebra
�
Solve linear equations in
one variable.
Here is a step-by-step procedure for solving a linear equation in one variable.
Not all of these steps are necessary to solve every equation.
Solving a Linear Equation
1. Simplify the algebraic expression on each side by removing grouping
symbols and combining like terms.
2. Collect all the variable terms on one side and all the numbers, or constant
terms, on the other side.
3. Isolate the variable and solve.
4. Check the proposed solution in the original equation.
Solving a Linear Equation
EXAMPLE 1
21x - 32 - 17 = 13 - 31x + 22.
Solve and check:
Solution
Step 1 Simplify the algebraic expression on each side.
Do not begin with 13 – 3. Multiplication
(the distributive property) is applied
before subtraction.
2(x-3)-17=13-3(x+2)
2x - 6 - 17 = 13 - 3x - 6
2x - 23 = - 3x + 7
This is the given equation.
Use the distributive property.
Combine like terms.
Step 2 Collect variable terms on one side and constant terms on the other
side. We will collect variable terms on the left by adding 3x to both sides. We will
collect the numbers on the right by adding 23 to both sides.
2x - 23 + 3x = - 3x + 7 + 3x
5x - 23 = 7
Add 3x to both sides.
Simplify: 2x + 3x = 5x.
5x - 23 + 23 = 7 + 23
5x = 30
Add 23 to both sides.
Simplify.
Step 3 Isolate the variable and solve. We isolate the variable, x, by dividing both
sides of 5x = 30 by 5.
5x
30
=
5
5
x = 6
Divide both sides by 5.
Simplify.
Step 4 Check the proposed solution in the original equation. Substitute 6 for x in
the original equation.
Discovery
Solve the equation in Example 1 by
collecting terms with the variable on
the right and numerical terms on the
left. What do you observe?
21x - 32 - 17 = 13 - 31x + 22
This is the original equation.
216 - 32 - 17 ⱨ 13 - 316 + 22
Substitute 6 for x.
2132 - 17 ⱨ 13 - 3182
Simplify inside parentheses.
6 - 17 ⱨ 13 - 24
Multiply.
-11 = - 11
Subtract.
The true statement - 11 = - 11 verifies that the solution set is 566.
Check Point
1
Solve and check:
412x + 12 = 29 + 312x - 52.
Section P.7 Equations 85
�
Solve linear equations containing
fractions.
Linear Equations with Fractions
Equations are easier to solve when they do not contain fractions. How do we
remove fractions from an equation? We begin by multiplying both sides of the
equation by the least common denominator of any fractions in the equation. The
least common denominator is the smallest number that all denominators will divide
into. Multiplying every term on both sides of the equation by the least common
denominator will eliminate the fractions in the equation. Example 2 shows how we
“clear an equation of fractions.”
Solving a Linear Equation Involving Fractions
EXAMPLE 2
x + 2
x - 1
= 2.
4
3
Solve and check:
Solution The fractional terms have denominators of 4 and 3. The smallest
number that is divisible by 4 and 3 is 12. We begin by multiplying both sides of the
equation by 12, the least common denominator.
x + 2
x - 1
= 2
4
3
x+2
x-1
12 a
b =12 ⴢ 2
4
3
This is the given equation.
Multiply both sides by 12.
12 a
x + 2
x - 1
b - 12 a
b = 24
4
3
Use the distributive property and
multiply each term on the left by 12.
12 ¢
4
x + 2
x - 1
≤ - 12 ¢
≤ = 24
4
3
Divide out common factors in each
multiplication on the left.
3
1
1
31x + 22 - 41x - 12 = 24
The fractions are now cleared.
3x + 6 - 4x + 4 = 24
Use the distributive property.
-x + 10 = 24
-x + 10 - 10 = 24 - 10
–x=14
Combine like terms: 3x - 4x = - x
and 6 + 4 = 10.
Subtract 10 from both sides.
Simplify.
We’re not finished. A
negative sign should not
precede the variable.
Isolate x by multiplying or dividing both sides of this equation by -1.
-x
14
=
-1
-1
x = - 14
Divide both sides by ⴚ 1.
Simplify.
Check the proposed solution. Substitute - 14 for x in the original equation.
You should obtain 2 = 2. This true statement verifies that the solution set is
5-146.
Check Point
2
Solve and check:
x - 3
5
x + 5
=
.
4
14
7
86 Chapter P Prerequisites: Fundamental Concepts of Algebra
�
Rational Equations
Solve rational equations with
variables in the denominators.
A rational equation is an equation containing one or more rational expressions. In
Example 2, we solved a rational equation with constants in the denominators. This
rational equation was a linear equation. Now, let’s consider a rational equation such as
3
1
4
+
= 2
.
x + 6
x - 2
x + 4x - 12
Can you see how this rational equation differs from the rational equation that we
solved earlier? The variable appears in the denominators. Although this rational
equation is not a linear equation, the solution procedure still involves multiplying
each side by the least common denominator. However, we must avoid any values of
the variable that make a denominator zero.
Solving a Rational Equation
EXAMPLE 3
Solve:
3
1
4
.
+
= 2
x + 6
x - 2
x + 4x - 12
Solution To identify values of x that make denominators zero, let’s factor
x2 + 4x - 12, the denominator on the right. This factorization is also necessary in
identifying the least common denominator.
1
4
3
+
=
x+6
x-2
(x+6)(x-2)
This denominator
is zero if x =฀−6.
This denominator
is zero if x =฀2.
This denominator is zero
if x =฀−6 or x =฀2.
We see that x cannot equal -6 or 2. The least common denominator is
1x + 621x - 22.
3
1
4
+
=
,฀ x Z - 6, x Z 2
x + 6
x - 2
1x + 621x - 22
(x+6)(x-2) a
1x + 62 1x - 22 #
3
1
4
+
b =(x+6)(x-2) ⴢ
x+6
x-2
(x+6)(x-2)
This is the given equation
with a denominator factored.
Multiply both sides by
1x + 621x - 22, the LCD.
Use the distributive property
3
1
4
+ 1x + 62 1x - 22 #
= 1x + 62 1x - 22 #
x + 6
x - 2
1x + 62 1x - 22 and divide out common
factors.
31x - 22 + 11x + 62 = 4
3x - 6 + x + 6 = 4
Simplify. This equation is
cleared of fractions.
Use the distributive property.
4x = 4
Combine like terms.
4x
4
=
4
4
Divide both sides by 4.
x = 1
Simplify. This is not part of
the restriction that x Z - 6
and x Z 2.
Check the proposed solution. Substitute 1 for x in the original equation. You should
obtain - 47 = - 47 . This true statement verifies that the solution set is 516.
Check Point
3
Solve:
6
5
-20
.
= 2
x + 3
x - 2
x + x - 6
Section P.7 Equations 87
Solving a Rational Equation
EXAMPLE 4
Solve:
1
1
2
= 2
.
x + 1
x - 1
x - 1
Solution We begin by factoring x2 - 1.
1
2
1
=
x+1
(x+1)(x-1)
x-1
This denominator
is zero if x = −1.
This denominator
is zero if x = −1 or x = 1.
This denominator
is zero if x = 1.
We see that x cannot equal - 1 or 1. The least common denominator is
1x + 121x - 12.
1
2
1
=
,฀ x Z - 1, x Z 1
x + 1
1x + 121x - 12
x - 1
This is the given equation
with a denominator factored.
(x+1)(x-1) ⴢ
1
2
1
=(x+1)(x-1) a
b
x+1
(x+1)(x-1)
x-1
1x + 12 1x - 12 #
1
2
1
and divide out common
= 1x + 12 1x - 12 #
- 1x + 12 1x - 12 #
x + 1
1x + 12 1x - 12
1x - 12 factors.
Use the distributive property
11x - 12 = 2 - 1x + 12
Simplify. This equation is
cleared of fractions.
x - 1 = 2 - x - 1
Simplify.
x - 1 = -x + 1
Combine numerical terms.
x + x - 1 = -x + x + 1
Add x to both sides.
2x - 1 = 1
Simplify.
2x - 1 + 1 = 1 + 1
Add 1 to both sides.
2x = 2
2x
2
=
2
2
x = 1
Study Tip
Reject any proposed solution that
causes any denominator in an equation to equal 0.
Simplify.
Divide both sides by 2.
Simplify.
The proposed solution, 1, is not a solution because of the restriction that x Z 1.
There is no solution to this equation. The solution set for this equation contains no
elements. The solution set is ⭋, the empty set.
Check Point
�
Multiply both sides by
1x + 121x - 12, the LCD.
Solve a formula for a variable.
4
Solve:
1
4
1
= 2
.
x + 2
x
2
x - 4
Solving a Formula for One of Its Variables
Solving a formula for a variable means rewriting the formula so that the variable is
isolated on one side of the equation. It does not mean obtaining a numerical value
for that variable.
To solve a formula for one of its variables, treat that variable as if it were the
only variable in the equation. Think of the other variables as if they were numbers.
88 Chapter P Prerequisites: Fundamental Concepts of Algebra
Solving a Formula for a Variable
EXAMPLE 5
If you wear glasses, did you know that each lens has a measurement called its focal
length, f? When an object is in focus, its distance from the lens, p, and the distance
from the lens to your retina, q, satisfy the formula
p
q
1
1
1
+
= .
p
q
f
Figure P.12
(See Figure P.12.) Solve this formula for p.
Solution Our goal is to isolate the variable p. We begin by multiplying both sides
by the least common denominator, pqf, to clear the equation of fractions.
1
1
1
+ =
q
p
f
We need to isolate p.
pqf a
p qfa
This is the given formula.
1
1
1
+ b =pqf a b
q
p
f
1
1
1
b + p q fa b = pq f a b
p
q
f
qf+pf=pq
Multiply both sides by pqf, the LCD.
Use the distributive property on the left side
and divide out common factors.
Simplify. The formula is cleared of
fractions.
We need to isolate p.
Study Tip
You cannot solve qf + pf = pq for
p by dividing both sides by q and
writing
qf + pf
= p.
q
To collect terms with p on one side of the equation, subtract pf from both sides. Then factor
p from the two resulting terms on the right to convert two occurrences of p into one.
qf + pf = pq
qf + pf - pf = pq - pf
When a formula is solved for a
specified variable, that variable must
be isolated on one side. The variable
p occurs on both sides of
Check Point
Factor out p, the specified variable.
Solve for q:฀
Divide both sides by q - f and solve for p.
Simplify.
1
1
1
+
= .
p
q
f
ƒ x ƒ = 2.
x
0
1
兩2兩 = 2
Figure P.13
qf = p1q - f2
We have seen that the absolute value of x, denoted ƒ x ƒ , describes the distance of x
from zero on a number line. Now consider an absolute value equation, such as
兩−2兩 = 2
−1
Simplify.
Equations Involving Absolute Value
Solve equations involving
absolute value.
−2
5
Subtract pf from both sides.
qf = pq - pf
p1q - f2
qf
=
q - f
q - f
qf
= p
q - f
qf + pf
= p.
q
�
This is the equation cleared of fractions.
2
This means that we must determine real numbers whose distance from the origin on
a number line is 2. Figure P.13 shows that there are two numbers such that ƒ x ƒ = 2,
namely, 2 and -2. We write x = 2 or x = - 2. This observation can be generalized as
follows:
Section P.7 Equations 89
Rewriting an Absolute Value Equation without Absolute Value Bars
If c is a positive real number and u represents any algebraic expression, then
ƒ u ƒ = c is equivalent to u = c or u = - c.
Solving an Equation Involving Absolute Value
EXAMPLE 6
Solve:
5 ƒ 1 - 4x ƒ - 15 = 0.
Solution
5|1-4x|-15=0
This is the given equation.
We need to isolate 兩1 − 4x兩,
the absolute value expression.
5 ƒ 1 - 4x ƒ = 15
ƒ 1 - 4x ƒ = 3
1 - 4x = 3฀ or฀ 1 - 4x = - 3
- 4x = 2 ฀ ฀ ฀ ฀
- 4x = - 4
1
x = - 2฀ ฀ ฀ ฀ ฀ ฀ ฀
x = 1
Add 15 to both sides.
Divide both sides by 5.
Rewrite ƒ u ƒ = c as u = c or u = - c.
Subtract 1 from both sides of each equation.
Divide both sides of each equation by -4.
Take a moment to check - 12 and 1, the proposed solutions, in the original equation,
5 ƒ 1 - 4x ƒ - 15 = 0. In each case, you should obtain the true statement 0 = 0. The
solution set is E - 12 , 1 F .
Check Point
6
Solve:
4 ƒ 1 - 2x ƒ - 20 = 0.
The absolute value of a number is never negative. Thus, if u is an algebraic
expression and c is a negative number, then ƒ u ƒ = c has no solution. For example,
the equation ƒ 3x - 6 ƒ = - 2 has no solution because ƒ 3x - 6 ƒ cannot be negative.
The solution set is ⭋, the empty set.
The absolute value of 0 is 0. Thus, if u is an algebraic expression and ƒ u ƒ = 0,
the solution is found by solving u = 0. For example, the solution of ƒ x - 2 ƒ = 0 is
obtained by solving x - 2 = 0. The solution is 2 and the solution set is 526.
�
Solve quadratic equations by
factoring.
Quadratic Equations and Factoring
Linear equations are first-degree polynomial equations of the form ax + b = 0.
Quadratic equations are second-degree polynomial equations and contain an
additional term involving the square of the variable.
Definition of a Quadratic Equation
A quadratic equation in x is an equation that can be written in the general form
ax2 + bx + c = 0,
where a, b, and c are real numbers, with a Z 0. A quadratic equation in x is also
called a second-degree polynomial equation in x.
Here are examples of quadratic equations in general form:
4x2-2x
a=4
b = −2
=0
c=0
2x2+7x-4=0.
a=2
b=7
c = −4
Some quadratic equations, including the two shown above, can be solved by
factoring and using the zero-product principle.
90 Chapter P Prerequisites: Fundamental Concepts of Algebra
The Zero-Product Principle
If the product of two algebraic expressions is zero, then at least one of the factors
is equal to zero.
If AB = 0, then A = 0 or B = 0.
The zero-product principle can be applied only when a quadratic equation is
in general form, with zero on one side of the equation.
Solving a Quadratic Equation by Factoring
1. If necessary, rewrite the equation in the general form ax2 + bx + c = 0,
moving all terms to one side, thereby obtaining zero on the other side.
2. Factor completely.
3. Apply the zero-product principle, setting each factor containing a variable
equal to zero.
4. Solve the equations in step 3.
5. Check the solutions in the original equation.
Solving Quadratic Equations by Factoring
EXAMPLE 7
Solve by factoring:
a. 4x2 - 2x = 0
b. 2x2 + 7x = 4.
Solution
4x2 - 2x = 0
a.
The given equation is in general form, with
zero on one side.
2x12x - 12 = 0
2x = 0
or
2x - 1 = 0
x = 0
2x = 1
1
x =
2
Factor.
Use the zero-product principle and set
each factor equal to zero.
Solve the resulting equations.
Check the proposed solutions, 0 and 12 , in the original equation.
Check 0:
4x2 - 2x = 0
4 # 02 - 2 # 0 ⱨ 0
0 - 0ⱨ0
0 = 0,
true
Check 12 :
4x2 - 2x = 0
4 A 12 B 2 - 2 A 12 B ⱨ 0
4 A 14 B - 2 A 12 B ⱨ 0
1 - 1ⱨ0
0 = 0,
The solution set is E 0, 12 F .
b.
2x2 + 7x = 4
2
2x + 7x - 4 = 4 - 4
This is the given equation.
Subtract 4 from both sides and write the
quadratic equation in general form.
2x2 + 7x - 4 = 0
12x - 121x + 42 = 0
2x - 1 = 0
or
x + 4 = 0
2x = 1
x = 12
฀฀฀฀฀฀
true
x = -4
Simplify.
Factor.
Use the zero-product principle and set each
factor equal to zero.
Solve the resulting equations.
Section P.7 Equations 91
Check the proposed solutions, 12 and -4, in the original equation.
Check 12 :
2x + 7x = 4
1 2
2A B + 7A1B ⱨ 4
Check -4:
2x2 + 7x = 4
21 - 422 + 71 -42 ⱨ 4
2
2
2
1
2
+
32 + 1 - 282 ⱨ 4
4 = 4, true
7
2
ⱨ4
4 = 4, true
The solution set is E -4, 12 F .
Check Point
7
Solve by factoring:
a. 3x2 - 9x = 0
�
Solve quadratic equations by the
square root property.
b. 2x2 + x = 1.
Quadratic Equations and the Square Root Property
Quadratic equations of the form u2 = d, where u is an algebraic expression and d is
a nonzero real number, can be solved by the square root property. First, isolate the
squared expression u2 on one side of the equation and the number d on the other
side. Then take the square root of both sides. Remember, there are two numbers
whose square is d. One number is 2d and one is - 2d.
We can use factoring to verify that u2 = d has these two solutions.
u2 = d
u2 - d = 0
This is the given equation.
Move all terms to one side and obtain
zero on the other side.
A u + 2d B A u - 2d B = 0
Factor.
u + 2d = 0
or฀ u - 2d = 0
u = 2d
u = - 2d
Set each factor equal to zero.
Solve the resulting equations.
Because the solutions differ only in sign, we can write them in abbreviated notation
as u = ; 2d. We read this as “u equals positive or negative square root of d” or
“u equals plus or minus square root of d.”
Now that we have verified these solutions, we can solve u2 = d directly by
taking square roots. This process is called the square root property.
The Square Root Property
If u is an algebraic expression and d is a positive real number, then u2 = d has
exactly two solutions:
If u2 = d, then u = 2d or u = - 2d.
Equivalently,
EXAMPLE 8
If u2 = d, then u = ; 2d.
Solving Quadratic Equations by the
Square Root Property
Solve by the square root property:
a. 3x2 - 15 = 0
b. 1x - 222 = 6.
Solution To apply the square root property, we need a squared expression by
itself on one side of the equation.
3x2-15=0
We want x2
by itself.
(x-2)2=6
The squared expression
is by itself.
92 Chapter P Prerequisites: Fundamental Concepts of Algebra
a.
3x2 - 15
3x2
x2
x = 25 or x
=
=
=
=
0
15
5
- 25
This is the original equation.
Add 15 to both sides.
Divide both sides by 3.
Apply the square root property.
Equivalently, x = ; 25.
By checking both proposed solutions in the original equation, we can confirm
that the solution set is E - 25, 25 F or E ; 25 F .
b. 1x - 222 = 6
x - 2 = ; 26
x = 2 ; 26
This is the original equation.
Apply the square root property.
Add 2 to both sides.
By checking both values in the original equation, we can confirm that the
solution set is E 2 + 26, 2 - 26 F or E 2 ; 26 F .
Check Point
8
Solve by the square root property:
b. 1x + 522 = 11.
a. 3x2 - 21 = 0
�
Solve quadratic equations by
completing the square.
Quadratic Equations and Completing the Square
How do we solve an equation in the form ax2 + bx + c = 0 if the trinomial
ax2 + bx + c cannot be factored? We cannot use the zero-product principle in such
a case. However, we can convert the equation into an equivalent equation that can be
solved using the square root property.This is accomplished by completing the square.
Completing the Square
b 2
If x2 + bx is a binomial, then by adding a b , which is the square of half the
2
coefficient of x, a perfect square trinomial will result. That is,
b 2
b 2
x2 + bx + a b = a x + b .
2
2
We can solve any quadratic equation by completing the square. If the coefficient of the x2-term is one, we add the square of half the coefficient of x to both
sides of the equation. When you add a constant term to one side of the equation to
complete the square, be certain to add the same constant to the other side of the
equation. These ideas are illustrated in Example 9.
EXAMPLE 9
Solving a Quadratic Equation
by Completing the Square
Solve by completing the square:
x2 - 6x + 4 = 0.
Solution We begin by subtracting 4 from both sides. This is done to isolate the
binomial x2 - 6x so that we can complete the square.
x2 - 6x + 4 = 0
x2 - 6x = - 4
2
This is the original equation.
Subtract 4 from both sides.
Next, we work with x - 6x = - 4 and complete the square. Find half the
coefficient of the x-term and square it. The coefficient of the x-term is - 6. Half of
- 6 is - 3 and 1-322 = 9. Thus, we add 9 to both sides of the equation.
Section P.7 Equations 93
x2 - 6x + 9 = - 4 + 9
1x - 322 = 5
x - 3 = 25
x = 3 + 25
or
Add 9 to both sides of x2 - 6x = - 4
to complete the square.
x - 3 = - 25
x = 3 - 25
Factor and simplify.
Apply the square root property.
Add 3 to both sides in each equation.
The solutions are 3 ; 25 and the solution set is E 3 + 25, 3 - 25 F , or E 3 ; 25 F .
Check Point
�
Solve quadratic equations using
the quadratic formula.
Deriving the Quadratic Formula
9
Solve by completing the square:
x2 + 4x - 1 = 0.
Quadratic Equations and the Quadratic Formula
We can use the method of completing the square to derive a formula that can be used
to solve all quadratic equations. The derivation given here also shows a particular
quadratic equation, 3x2 - 2x - 4 = 0, to specifically illustrate each of the steps.
Notice that if the coefficient of the x2-term in a quadratic equation is not one, you
must divide each side of the equation by this coefficient before completing the square.
General Form
of a Quadratic Equation
Comment
A Specific Example
ax2 + bx + c = 0, a 7 0
This is the given equation.
3x2 - 2x - 4 = 0
x2 +
b
c
= 0
x +
a
a
Divide both sides by a so that the coefficient of x2 is 1.
x2 +
b
c
x = a
a
b
c
b 2
b 2
x2+ x + a b = - + a b
a
2a
a
2a
x2 -
2
4
x - = 0
3
3
c
Isolate the binomial by adding - on both sides of
a
the equation.
x2 -
4
2
x =
3
3
Complete the square. Add the square of half the
coefficient of x to both sides.
1 2 4
1 2
2
x2- x + a- b = + a- b
3
3
3
3
(half)2
b2
b
b2
c
+
x2 + x +
=
a
a
4a2
4a2
ax +
c 4a
b 2
b2
b = - #
+
a 4a
2a
4a2
-4ac + b2
b 2
b =
2a
4a2
b 2
b2 - 4ac
ax +
b =
2a
4a2
ax +
x +
x +
x =
x =
b
b2 - 4ac
= ;
2a
B 4a2
b
2b2 - 4ac
= ;
2a
2a
-b
3b2 - 4ac
;
2a
2a
- b ; 3b2 - 4ac
2a
(half)2
2
1
4
1
x2 - x + = +
3
9
3
9
4 3
1 2
1
b = # +
3
3 3
9
Factor on the left side and obtain a common
denominator on the right side.
ax -
Add fractions on the right side.
1 2
12 + 1
b =
3
9
1 2
13
ax - b =
3
9
Apply the square root property.
Take the square root of the quotient, simplifying the
denominator.
Solve for x by subtracting
b
from both sides.
2a
Combine fractions on the right side.
ax -
x -
x -
x =
x =
1
13
= ;
3
A9
1
213
= ;
3
3
1
213
;
3
3
1 ; 213
3
The formula shown at the bottom of the left column is called the quadratic
formula. A similar proof shows that the same formula can be used to solve quadratic
equations if a, the coefficient of the x2-term, is negative.
94 Chapter P Prerequisites: Fundamental Concepts of Algebra
The Quadratic Formula
The solutions of a quadratic equation in general form ax2 + bx + c = 0, with
a Z 0, are given by the quadratic formula
x=
–b_兹b2-4ac
.
2a
x equals negative b plus or minus
the square root of b2 − 4ac, all
divided by 2a.
To use the quadratic formula, write the quadratic equation in general form if
necessary. Then determine the numerical values for a (the coefficient of the x2-term),
b (the coefficient of the x-term), and c (the constant term). Substitute the values of
a, b, and c into the quadratic formula and evaluate the expression. The ; sign
indicates that there are two solutions of the equation.
EXAMPLE 10 Solving a Quadratic Equation Using
the Quadratic Formula
Solve using the quadratic formula:
2x2 - 6x + 1 = 0.
Solution The given equation is in general form. Begin by identifying the values
for a, b, and c.
2x2-6x+1=0
a=2
b = −6
c=1
Substituting these values into the quadratic formula and simplifying gives the
equation’s solutions.
x =
=
=
Study Tip
Checking irrational solutions can be
time-consuming. The solutions given
by the quadratic formula are always
correct, unless you have made a
careless
error.
Checking
for
computational errors or errors in
simplification is sufficient.
=
=
=
-b ; 3b2 - 4ac
2a
Use the quadratic formula.
- 1- 62 ; 41- 622 - 4122112
2#2
6 ; 236 - 8
4
6 ; 228
4
- 1- 62 = 6, 1 - 622 = 1- 621 -62 = 36, and
4122112 = 8.
Complete the subtraction under the radical.
6 ; 2 27
4
228 = 24 # 7 = 2427 = 2 27
2 A 3 ; 27 B
Factor out 2 from the numerator.
4
3 ; 27
=
2
Divide the numerator and denominator by 2.
The solution set is b
Check Point
Substitute the values for a, b, and c:
a = 2, b = - 6, and c = 1.
10
3 + 27 3 - 27
3 ; 27
,
r or b
r.
2
2
2
Solve using the quadratic formula:
2x2 + 2x - 1 = 0.
Section P.7 Equations 95
쐅
Use the discriminant to
determine the number and type
of solutions of quadratic
equations.
Quadratic Equations and the Discriminant
The quantity b2 - 4ac, which appears under the radical sign in the quadratic
formula, is called the discriminant. Table P.4 shows how the discriminant of the
quadratic equation ax2 + bx + c = 0 determines the number and type of solutions.
Table P.4 The Discriminant and the Kinds of Solutions to ax2 ⴙ bx ⴙ c ⴝ 0
Discriminant b2 ⴚ 4ac
Kinds of Solutions to ax2 ⴙ bx ⴙ c ⴝ 0
b2 - 4ac 7 0
Two unequal real solutions: If a, b, and c are rational
numbers and the discriminant is a perfect square, the
solutions are rational. If the discriminant is not a perfect
square, the solutions are irrational.
b2 - 4ac = 0
One solution (a repeated solution) that is a real number: If
a, b, and c are rational numbers, the repeated solution is also
a rational number.
b2 - 4ac 6 0
No real solutions
EXAMPLE 11 Using the Discriminant
Compute the discriminant of 4x2 - 8x + 1 = 0. What does the discriminant
indicate about the number and type of solutions?
Solution Begin by identifying the values for a, b, and c.
2x2-6x+1=0
a=2
b = −6
c=1
Substitute and compute the discriminant:
b2 - 4ac = 1-822 - 4 # 4 # 1 = 64 - 16 = 48.
The discriminant is 48. Because the discriminant is positive, the equation
4x2 - 8x + 1 = 0 has two unequal real solutions.
11
Compute the discriminant of 3x2 - 2x + 5 = 0. What does
the discriminant indicate about the number and type of solutions?
Check Point
쐈
Determine the most efficient
method to use when solving a
quadratic equation.
Determining Which Method to Use
All quadratic equations can be solved by the quadratic formula. However, if an
equation is in the form u2 = d, such as x2 = 5 or 12x + 322 = 8, it is faster to use
the square root property, taking the square root of both sides. If the equation is not
in the form u2 = d, write the quadratic equation in general form
1ax2 + bx + c = 02. Try to solve the equation by factoring. If ax2 + bx + c cannot
be factored, then solve the quadratic equation by the quadratic formula.
Because we used the method of completing the square to derive the quadratic
formula, we no longer need it for solving quadratic equations. However, we will
use completing the square later in the book to help graph circles and other kinds
of equations.
Table P.5 on the next page summarizes our observations about which
technique to use when solving a quadratic equation.
96 Chapter P Prerequisites: Fundamental Concepts of Algebra
Table P.5 Determining the Most Efficient Technique to Use When Solving a Quadratic Equation
Description and Form
of the Quadratic Equation
Most Efficient
Solution Method
ax2 + bx + c = 0 and ax2 + bx + c
can be factored easily.
Factor and use the
zero-product principle.
3x2 + 5x - 2 = 0
13x - 121x + 22 = 0
or
x + 2 = 0
3x - 1 = 0
1
฀฀฀฀ x =
฀฀
x = -2
3
ax2 + bx = 0
The quadratic equation has no
constant term. 1c = 02
Factor and use the
zero-product principle.
6x2 + 9x = 0
3x12x + 32 = 0
or
2x + 3 = 0
3x = 0
฀
2x = - 3
x = 0฀ ฀ ฀
x = - 23
ax2 + c = 0
The quadratic equation has
no x-term. 1b = 02
Solve for x2 and apply the
square root property.
u2 = d; u is a first-degree polynomial.
Example
7x2 - 4 = 0
7x2 = 4
4
x2 =
7
x = ;
4
A7
= ;
27
2
= ;
# 27
27 27
2
= ;
227
7
1x + 422 = 5
Use the square root property.
x + 4 = ; 25
x = - 4 ; 25
ax2 + bx + c = 0 and ax2 + bx + c cannot
be factored or the factoring is too difficult.
Use the quadratic formula:
x¤-2x-6=0
-b ; 3b - 4ac
.
2a
2
x =
a=1
x =
=
b = −2
c = −6
-1-22 ; 41- 222 - 41121 - 62
2(1)
2 ; 44 + 24
2(1)
2 ; 228
2 ;
=
2
2A1
2 ; 227
=
=
2
=
= 1 ; 27
쐉
Solve radical equations.
24 27
2
; 27 B
2
Radical Equations
A radical equation is an equation in which the variable occurs in a square root, cube
root, or any higher root. An example of a radical equation is
1x = 9.
We solve the equation by squaring both sides:
Squaring both
sides eliminates
the square root.
A兹xB =92
2
x=81.
The proposed solution, 81, can be checked in the original equation, 1x = 9.
Because 281 = 9, the solution is 81 and the solution set is 5816.
Section P.7 Equations
97
In general, we solve radical equations with square roots by squaring both sides
of the equation. We solve radical equations with nth roots by raising both sides of
the equation to the nth power. Unfortunately, if n is even, all the solutions of the
equation raised to the even power may not be solutions of the original equation.
Consider, for example, the equation
x = 4.
If we square both sides, we obtain
x2 = 16.
Solving this equation using the square root property, we obtain
x = ; 216 = ; 4.
The new equation x = 16 has two solutions, - 4 and 4. By contrast, only 4 is a solution
of the original equation, x = 4. For this reason, when raising both sides of an equation
to an even power, always check proposed solutions in the original equation.
Here is a general method for solving radical equations with nth roots:
2
Solving Radical Equations Containing nth Roots
1. If necessary, arrange terms so that one radical is isolated on one side of the
equation.
2. Raise both sides of the equation to the nth power to eliminate the isolated
nth root.
3. Solve the resulting equation. If this equation still contains radicals, repeat
steps 1 and 2.
4. Check all proposed solutions in the original equation.
Extra solutions may be introduced when you raise both sides of a radical
equation to an even power. Such solutions, which are not solutions of the given
equation, are called extraneous solutions or extraneous roots.
EXAMPLE 12 Solving a Radical Equation
Solve: 22x - 1 + 2 = x.
Solution
Step 1 Isolate a radical on one side. We isolate the radical, 22x - 1, by
subtracting 2 from both sides.
22x - 1 + 2 = x
22x - 1 = x - 2
Study Tip
Be sure to square both sides of an
equation. Do not square each term.
Correct:
A 22x - 1 B 2 = 1x - 222
Incorrect!
A 22x - 1 B 2 = x2 - 2 2
This is the given equation.
Subtract 2 from both sides.
Step 2 Raise both sides to the nth power. Because n, the index, is 2, we square
both sides.
A 22x - 1 B 2 = 1x - 222
2x - 1 = x2 - 4x + 4
Simplify. Use the formula
1A - B22 = A2 - 2AB + B2
on the right side.
Step 3 Solve the resulting equation. Because of the x2-term, the resulting equation
is a quadratic equation. We can obtain 0 on the left side by subtracting 2x and
adding 1 on both sides.
2x - 1 = x2 - 4x + 4
The resulting equation is quadratic.
0 = x2 - 6x + 5
0 = 1x - 121x - 52
x - 1 = 0฀ or฀ x - 5 = 0
x = 1
x = 5
Write in general form, subtracting 2x and
adding 1 on both sides.
Factor.
Set each factor equal to 0.
Solve the resulting equations.
98 Chapter P Prerequisites: Fundamental Concepts of Algebra
Step 4 Check the proposed solutions in the original equation.
Check 1:
Check 5:
22x - 1 + 2 = x
22x - 1 + 2 = x
22 # 1 - 1 + 2 ⱨ 1
22 # 5 - 1 + 2 ⱨ 5
21 + 2 ⱨ 1
1 + 2ⱨ1
29 + 2 ⱨ 5
3 + 2ⱨ5
3 = 1,
5 = 5,
false
true
Thus, 1 is an extraneous solution.The only solution is 5, and the solution set is 556.
Check Point
12
Solve:
2x + 3 + 3 = x.
Exercise Set P.7
Practice Exercises
25.
5
6
1
= 2
x - 4
x + 2
x - 2x - 8
26.
2
8
1
= 2
x - 3
x + 1
x - 2x - 3
In Exercises 1–16, solve each linear equation.
1. 7x - 5 = 72
2. 6x - 3 = 63
3. 11x - 16x - 52 = 40
4. 5x - 12x - 102 = 35
5. 2x - 7 = 6 + x
6. 3x + 5 = 2x + 13
7. 7x + 4 = x + 16
8. 13x + 14 = 12x - 5
9. 31x - 22 + 7 = 21x + 52
x + 1
1
2 - x
= +
4
6
3
11.
x + 3
3
x - 5
= +
6
8
4
12.
13.
x
x - 3
= 2 +
4
3
14. 5 +
x - 2
x + 3
=
3
8
x + 1
x + 2
x - 3
x + 2
3x
= 5 =
16.
3
7
5
2
3
Exercises 17–26 contain rational equations with variables in
denominators. For each equation, a. Write the value or values of
the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the
equation.
15.
19.
8x
8
= 4 x + 1
x + 1
21.
3
1
2
+ =
2x - 2
2
x - 1
22.
23.
3
-4
- 7 =
18.
x + 4
x + 4
20.
27. I = Prt for P
28. C = 2pr for r
29. T = D + pm for p
30. P = C + MC for M
1
2 h1a
32. A = 21 h1a + b2 for b
31. A =
10. 21x - 12 + 3 = x - 31x + 12
1
11
+ 5 =
17.
x - 1
x - 1
In Exercises 27–42, solve each formula for the specified variable.
Do you recognize the formula? If so, what does it describe?
2
x
=
- 2
x - 2
x - 2
+ b2 for a
33. S = P + Prt for r
F
35. B =
for S
S - V
34. S = P + Prt for t
C
36. S =
for r
1 - r
37. IR + Ir = E for I
1
1
1
39.
+
= for f
p
q
f
f1f2
41. f =
for f1
f1 + f2
38. A = 2lw + 2lh + 2wh for h
1
1
1
40.
for R1
=
+
R
R1
R2
f1f2
42. f =
for f2
f1 + f2
In Exercises 43–54, solve each absolute value equation or indicate
the equation has no solution.
43. ƒ x - 2 ƒ = 7
44. ƒ x + 1 ƒ = 5
45. ƒ 2x - 1 ƒ = 5
46. ƒ 2x - 3 ƒ = 11
47. 2 ƒ 3x - 2 ƒ = 14
48. 3 ƒ 2x - 1 ƒ = 21
49. 2 ` 4 -
50. 4 ` 1 -
5
x ` + 6 = 18
2
3
x ` + 7 = 10
4
51. ƒ x + 1 ƒ + 5 = 3
52. ƒ x + 1 ƒ + 6 = 2
5
1
3
=
+
x + 3
2x + 6
x - 2
53. ƒ 2x - 1 ƒ + 3 = 3
54. ƒ 3x - 2 ƒ + 4 = 4
1
2x
2
= 2
x + 1
x - 1
x - 1
55. x2 - 3x - 10 = 0
56. x2 - 13x + 36 = 0
57. x2 = 8x - 15
58. x2 = - 11x - 10
59. 5x2 = 20x
60. 3x2 = 12x
2
32
4
+
= 2
24.
x + 5
x - 5
x - 25
In Exercises 55–60, solve each quadratic equation by factoring.
Section P.7 Equations 99
In Exercises 61–66, solve each quadratic equation by the square
root property.
61. 3x2 = 27
62. 5x2 = 45
63. 5x2 + 1 = 51
64. 3x2 - 1 = 47
65. 31x - 422 = 15
66. 31x + 422 = 21
In Exercises 67–74, solve each quadratic equation by completing
the square.
2
67. x + 6x = 7
2
69. x - 2x = 2
2
2
68. x + 6x = - 8
2
70. x + 4x = 12
2
71. x - 6x - 11 = 0
72. x - 2x - 5 = 0
73. x2 + 4x + 1 = 0
74. x2 + 6x - 5 = 0
In Exercises 75–82, solve each quadratic equation using the
quadratic formula.
2
75. x + 8x + 15 = 0
2
77. x + 5x + 3 = 0
2
76. x + 8x + 12 = 0
2
78. x + 5x + 2 = 0
2
80. 5x + x - 2 = 0
2
82. 3x2 = 6x - 1
79. 3x - 3x - 4 = 0
81. 4x = 2x + 7
2
Compute the discriminant of each equation in Exercises 83–90.
What does the discriminant indicate about the number and type of
solutions?
83. x2 - 4x - 5 = 0
84. 4x2 - 2x + 3 = 0
85. 2x2 - 11x + 3 = 0
86. 2x2 + 11x - 6 = 0
2
2
87. x = 2x - 1
88. 3x = 2x - 1
89. x2 - 3x - 7 = 0
90. 3x2 + 4x - 2 = 0
In Exercises 91–114, solve each quadratic equation by the method
of your choice.
2
91. 2x - x = 1
2
92. 3x - 4x = 4
2
94. 5x = 6 - 13x
2
95. 3x = 60
96. 2x2 = 250
97. x2 - 2x = 1
98. 2x2 + 3x = 1
93. 5x + 2 = 11x
99. 12x + 321x + 42 = 1
101. 13x - 42 = 16
2
2
2
100. 12x - 521x + 12 = 2
117. 2x + 3 = x - 3
118. 2x + 10 = x - 2
121. x - 22x + 5 = 5
122. x - 2x + 11 = 1
119. 22x + 13 = x + 7
123. 22x + 19 - 8 = x
In Exercises 125–134, solve each equation.
125. 25 - 32 + 5x - 31x + 224 =
-312x - 52 - 351x - 12 - 3x + 34
126. 45 - 34 - 2x - 41x + 724 =
-411 + 3x2 - 34 - 31x + 22 - 212x - 524
127. 7 - 7x = 13x + 221x - 12
128. 10x - 1 = 12x + 122
129. ƒ x2 + 2x - 36 ƒ = 12
130. ƒ x2 + 6x + 1 ƒ = 8
1
1
5
131. 2
=
+ 2
x + 2
x - 3x + 2
x - 4
x
1
x - 1
132.
+
= 2
x - 2
x - 3
x - 5x + 6
133. 2x + 8 - 2x - 4 = 2
134. 2x + 5 - 2x - 3 = 2
In Exercises 135–136, list all numbers that must be excluded
from the domain of each rational expression.
7
3
135.
136.
2x2 + 4x - 9
2x2 - 8x + 5
Application Exercises
The latest guidelines, which apply to both men and women, give
healthy weight ranges, rather than specific weights, for your height.
The further you are above the upper limit of your range, the
greater are the risks of developing weight-related health problems.
The bar graph shows these ranges for various heights for people
between the ages of 19 and 34, inclusive.
102. 12x + 722 = 25
Healthy Weight Ranges for
Men and Women, Ages 19 to 34
2
104. 9 - 6x + x = 0
105. 4x2 - 16 = 0
106. 3x2 - 27 = 0
220
107. x2 = 4x - 2
108. x2 = 6x - 7
200
109. 2x2 - 7x = 0
110. 2x2 + 5x = 3
180
112.
1
1
1
=
+
x
x + 3
4
6
28
2x
+
= - 2
x - 3
x + 3
x - 9
3
5
x2 - 20
114.
+
= 2
x - 3
x - 4
x - 7x + 12
113.
In Exercises 115–124, solve each radical equation. Check all
proposed solutions.
115. 23x + 18 = x
116. 220 - 8x = x
Lower end of range
Weight (pounds)
1
1
1
=
+
x
x + 2
3
124. 22x + 15 - 6 = x
Practice Plus
103. 3x - 12x + 12 = 0
111.
120. 26x + 1 = x - 1
160
146
120
111
118
125
132
195
184
174
164
155
140
Upper end of range
140
148
100
80
60
5⬘4⬙
5⬘6⬙
5⬘8⬙
5⬘10⬙
Height
Source: U.S. Department of Health and Human Services
6⬘0⬙
6⬘2⬙
100 Chapter P Prerequisites: Fundamental Concepts of Algebra
The mathematical model
Use the formula at the bottom of the previous column to solve
Exercises 141–142.
How many liters of pure peroxide should be added to
produce a new product that is 28% peroxide?
140. Suppose that x liters of pure acid are added to 200 liters of a
35% acid solution.
a. Write a formula that gives the concentration, C, of the
new mixture. (Hint: See Exercise 139.)
b. How many liters of pure acid should be added to
produce a new mixture that is 74% acid?
A driver’s age has something to do with his or her chance of getting
into a fatal car crash. The bar graph shows the number of fatal
vehicle crashes per 100 million miles driven for drivers of various
age groups. For example, 25-year-old drivers are involved in
4.1 fatal crashes per 100 million miles driven. Thus, when a group
of 25-year-old Americans have driven a total of 100 million miles,
approximately 4 have been in accidents in which someone died.
Fatal Crashes per 100 Million
Miles Driven
Age of U.S. Drivers and Fatal Crashes
18
17.7
16.3
15
12
9.5
9
8.0
6.2
6
4.1
3
0
16
18
20
25
2.8
2.4
3.0
35
45
55
Age of Drivers
3.8
65
75
79
Source: Insurance Institute for Highway Safety
The number of fatal vehicle crashes per 100 million miles, y, for
drivers of age x can be modeled by the formula
y = 0.013x2 - 1.19x + 28.24.
Average Nonprogram
Minutes in an Hour of
Prime-Time Cable TV
16
15
14.3
14
13
14.6
13.5
12.5
12
11
10
2005
139. A company wants to increase the 10% peroxide content
of its product by adding pure peroxide (100% peroxide).
If x liters of pure peroxide are added to 500 liters of its
10% solution, the concentration, C, of the new mixture is
given by
x + 0.115002
C =
.
x + 500
By 2005, the amount of “clutter,”
including commercials and plugs
for other shows, had increased to
the point where an “hour-long”
drama on cable TV was 45.4 minutes. The bar graph shows the
average number of nonprogram
minutes in an hour of prime-time
cable television.
2002
138. Use the formula to find a healthy weight for a person whose
height is 6¿0–. (Hint: H = 12 because this person’s height is
12 inches over 5 feet.) How many pounds is this healthy
weight below the upper end of the range shown by the bar
graph on the previous page?
142. What age groups are expected to be involved in 10 fatal
crashes per 100 million miles driven? How well does the
formula model the trend in the actual data shown by the bar
graph in the previous column?
1999
137. Use the formula to find a healthy weight for a person
whose height is 5¿6–. (Hint: H = 6 because this person’s
height is 6 inches over 5 feet.) How many pounds is this
healthy weight below the upper end of the range shown by
the bar graph on the previous page?
1996
describes a weight, W, in pounds, that lies within the healthy weight
range shown by the bar graph on the previous page for a person
whose height is H inches over 5 feet. Use this information to solve
Exercises 137–138.
141. What age groups are expected to be involved in 3 fatal crashes
per 100 million miles driven? How well does the formula
model the trend in the actual data shown by the bar graph in
the previous column?
Non-Program Minutes
in an Hour
W
- 3H = 53
2
Year
Source: Nielsen Monitor-Plus
M = 0.7 1x + 12.5,
The data can be modeled by the formula
where M is the average number of nonprogram minutes in an
hour of prime-time cable x years after 1996. Use the formula to
solve Exercises 143–144.
143. Assuming the trend from 1996 through 2005 continues, use
the model to project when there will be 15.1 cluttered
minutes in every prime-time cable TV hour.
144. Assuming the trend from 1996 through 2005 continues, use
the model to project when there will be 16 cluttered
minutes in every prime-time cable TV hour.
Writing in Mathematics
145. What is a linear equation in one variable? Give an example
of this type of equation.
146. Explain how to determine the restrictions on the variable
for the equation
3
4
7
.
+
= 2
x + 5
x - 2
x + 3x - 6
147. What does it mean to solve a formula for a variable?
148. Explain how to solve an equation involving absolute value.
149. Why does the procedure that you explained in Exercise 148
not apply to the equation ƒ x - 2 ƒ = - 3? What is the
solution set for this equation?
150. What is a quadratic equation?
151. Explain how to solve x2 + 6x + 8 = 0 using factoring and
the zero-product principle.
152. Explain how to solve x2 + 6x + 8 = 0 by completing the
square.
153. Explain how to solve x2 + 6x + 8 = 0 using the quadratic
formula.
154. How is the quadratic formula derived?
155. What is the discriminant and what information does it
provide about a quadratic equation?
Section P.8 Modeling with Equations 101
156. If you are given a quadratic equation, how do you
determine which method to use to solve it?
157. In solving 22x - 1 + 2 = x, why is it a good idea to
isolate the radical term? What if we don’t do this and simply
square each side? Describe what happens.
158. What is an extraneous solution to a radical equation?
Critical Thinking Exercises
Make Sense? In Exercises 159–162, determine whether each
statement makes sense or does not make sense, and explain your
reasoning.
159. The model P = - 0.18n + 2.1 describes the number of pay
phones, P, in millions, n years after 2000, so I have to solve a
linear equation to determine the number of pay phones in
2006.
160. Although I can solve 3x + 15 = 41 by first subtracting 51 from
both sides, I find it easier to begin by multiplying both sides
by 20, the least common denominator.
161. Because I want to solve 25x2 - 169 = 0 fairly quickly, I’ll
use the quadratic formula.
162. When checking a radical equation’s proposed solution, I can
substitute into the original equation or any equation that is
part of the solution process.
In Exercises 163–166, determine whether each statement is true or
false. If the statement is false, make the necessary change(s) to
produce a true statement.
163. The equation 12x - 322 = 25 is equivalent to 2x - 3 = 5.
Section
P.8
164. Every quadratic equation has two distinct numbers in its
solution set.
165. The equations 3y - 1 = 11 and 3y - 7 = 5 are
equivalent.
166. The equation ax2 + c = 0, a Z 0, cannot be solved by the
quadratic formula.
7x + 4
167. Find b such that
+ 13 = x will have a solution set
b
given by 5 - 66.
168. Write a quadratic equation in general form whose solution
set is 5 - 3, 56.
C - S
N.
169. Solve for C:฀ V = C L
2
170. Solve for t:฀ s = - 16t + v0t.
Preview Exercises
Exercises 171–173 will help you prepare for the material covered
in the next section.
171. Jane’s salary exceeds Jim’s by $150 per week. If x represents
Jim’s weekly salary, write an algebraic expression that
models Jane’s weekly salary.
172. A long-distance telephone plan has a monthly fee of $20
with a charge of $0.05 per minute for all long-distance calls.
Write an algebraic expression that models the plan’s
monthly cost for x minutes of long-distance calls.
173. If the width of a rectangle is represented by x and the length
is represented by x + 200, write a simplified algebraic
expression that models the rectangle’s perimeter.
Modeling with Equations
Objective
How Long It Takes to Earn $1000
� Use equations to solve
problems.
Howard Stern
Radio host
24 sec.
Dr. Phil McGraw
Television host
2 min. 24 sec.
Brad Pitt
Actor
4 min. 48 sec.
Kobe Bryant
Basketball player
5 min. 30 sec.
Chief executive
U.S. average
2 hr. 55 min.
Doctor, G.P.
U.S. average
13 hr. 5 min.
High school teacher
U.S. average
43 hours
Janitor
U.S. average
103 hours
Source: Time
I
n this section, you’ll see examples and exercises focused on how much money
Americans earn. These situations illustrate a step-by-step strategy for solving
problems. As you become familiar with this strategy, you will learn to solve a wide
variety of problems.